Distance 12 X Acceleration X (time) ^2

The vector acceleration A has two components A x and A y given by: (acceleration along the y axis only) A x = 0 and A y = - g = - 9.8 m/s 2 At time t, the velocity has two components given by V x = V 0 cos(θ) and V y = V 0 sin(θ) - g t The displacement is a vector with the components x and y given by: x = V 0 cos(θ) t and y = y 0 + V 0 sin. You end up with time squared in the denominator because you divide velocity by time. In other words, acceleration is the rate at which your velocity changes, because rates have time in the denominator. For acceleration, you see units of meters per second 2, centimeters per second 2, miles per second 2, feet per second 2, or even kilometers per. About Speed Distance Time Calculator. This online calculator tool can be a great help for calculating time basing on such physical concepts as speed and distance. Therefore, in order to calculate the time, both distance and speed parameters must be entered. For the speed, you need to enter its value and select speed unit by using the scroll down menu in the calculator. Distance = (1/2 of acceleration) x (time squared)You can change this around to solve it for acceleration or time.(Time squared) = (distance)/(half of acceleration)Time = the square root of (2 x.

1. Distance 12 X Acceleration X (time) ^2tion X Time 2
2. Distance Using Acceleration And Time
3. Finding Distance With Acceleration And Time
4. Given Distance And Time Find Acceleration

Basic Relationships

Recall from our study of derivatives that if (xleft( t right)) is the position of an object moving along a straight line at time (t,) then the velocity of the object is

[vleft( t right) = frac{{dx}}{{dt}},]

and the acceleration is given by

[aleft( t right) = frac{{dv}}{{dt}} = frac{{{d^2}x}}{{d{t^2}}}.]

Using the integral calculus, we can calculate the velocity function from the acceleration function, and the position function from the velocity function.

The Fundamental Theorem of Calculus says that

[{intlimits_{{t_1}}^{{t_2}} {aleft( t right)dt} = left. {vleft( t right)} right|_{{t_1}}^{{t_2}} }={ vleft( {{t_2}} right) – vleft( {{t_1}} right).}]

Similarly, the difference between the position at time ({{t_1}}) and the position at time ({{t_2}}) is determined by the equation

[{intlimits_{{t_1}}^{{t_2}} {vleft( t right)dt} = left. Big ben slot machine free play. {xleft( t right)} right|_{{t_1}}^{{t_2}} }={ xleft( {{t_2}} right) – xleft( {{t_1}} right).}]

If the object moves from the position (xleft( {{t_1}} right)) to the position (xleft( {{t_2}} right),) the change (xleft( {{t_2}} right) – xleft( {{t_1}} right)) is called the displacement of the object:

[Delta x = xleft( {{t_2}} right) – xleft( {{t_1}} right).]

To find the total distance traveled by the object between time ({{t_1}}) and time ({{t_2}},) we need to compute the integral of (left| {vleft( t right)} right|:)

[d = intlimits_{{t_1}}^{{t_2}} {left| {vleft( t right)} right|dt} .]

Constant Acceleration

Suppose that an object is moving along a straight line with the constant acceleration (a.) At time ({t_1} = 0,) the object has an initial velocity ({v_0}) and an initial position ({x_0}.)

The velocity and position of the object at time (t) are given by the equations

[vleft( t right) = {v_0}t + at,]

[xleft( t right) = {x_0} + {v_0}t + frac{{a{t^2}}}{2}.]

Solved Problems

Click or tap a problem to see the solution.

Distance 12 X Acceleration X (time) ^2tion X Time 2

Solution.

Given that the velocity is positive for (t gt 0,) the total distance traveled for the time interval (left[ {0,t} right)) is expressed by the integral:

[{dleft( t right) = intlimits_0^t {left| {vleft( u right)} right|du} }={ intlimits_0^t {vleft( u right)du} }={ intlimits_0^t {sqrt {4 +u} du} ,}]

where (u) is the inner variable which has no impact on the computation of the integral.

Integration yields:

[{dleft( t right) = left. {frac{{2sqrt {{{left( {4 + u} right)}^3}} }}{3}} right|_0^t }={ frac{2}{3}left[ {sqrt {{{left( {4 + t} right)}^3}} – 8} right].}]

Substituting (t = 5,) we have Screens 3 6 12 – access your computer remotely.

[{dleft( {t = 5,text{s}} right) = frac{2}{3}left[ {sqrt {{{left( {4 + 5} right)}^3}} – 8} right] }={ frac{2}{3} cdot 19 }={ frac{{38}}{3} approx 12.7,text{m}}]

Solution.

The equation of motion of the particle has the form

[v = frac{{dx}}{{dt}} = 2sqrt x .]

We have a simple differential equation that describes the particle's position as a function of time. Separating the variables and integrating both sides yields:

[{frac{{dx}}{{sqrt x }} = 2dt,};; Rightarrow {int {frac{{dx}}{{sqrt x }}} = 2int {dt} ,};; Rightarrow {2sqrt x = 2t + C,};; Rightarrow {sqrt x = t + C.}]

It follows from the initial condition (xleft( {t = 0} right) = 0) that the (C = 0.) Hence, the particle moves according to the law:

[x = {t^2}.]

It is easy to see that (t = 10,text{s}) when (x = 100,text{m}.)

Solution.

Let's first compute the displacement (Delta x) of the object when (t = 5,text{s}.) Integrating the velocity expression, we obtain

[{Delta x = intlimits_{{t_1}}^{{t_2}} {vleft( t right)dt} }={ intlimits_0^5 {left( {6 – 2t} right)dt} }={ left. {left( {6t – {t^2}} right)} right|_0^5 }={ 5,text{m}.}]

Notice that the velocity changes sign at ({t_1} = 3.) Therefore, to calculate the distance (d) traveled by the object we split the initial interval (left[ {0,5} right]) into two subintervals (left[ {0,3} right]) and (left[ {3,5} right].) This yields

[{d = intlimits_{{t_1}}^{{t_2}} {left| {vleft( t right)} right|dt} }={ intlimits_0^5 {left| {6 – 2t} right|dt} }={ intlimits_0^3 {left| {6 – 2t} right|dt} + intlimits_3^5 {left| {6 – 2t} right|dt} }={ intlimits_0^3 {left( {6 – 2t} right)dt} – intlimits_3^5 {left( {6 – 2t} right)dt} }={ left. {left( {6t – {t^2}} right)} right|_0^3 – left. {left( {6t – {t^2}} right)} right|_3^5 }={ left[ {left( {18 – 9} right) – 0} right] }-{ left[ {left( {30 – 25} right) – left( {18 – 9} right)} right] }={ 9 + 4 }={ 13,text{m}.}]

Solution.

Given that the initial velocity is zero: ({v_0} = 0,) we determine the velocity equation:

[{vleft( t right) = int {aleft( t right)dt} = int {cos frac{{pi t}}{6}dt} }={ frac{6}{pi }sin frac{{pi t}}{6}.}]

The distance traveled in the (3text{rd}) second is

[{d = intlimits_2^3 {vleft( t right)dt} }={ intlimits_2^3 {frac{6}{pi }sin frac{{pi t}}{6}dt} }={ frac{{{6^2}}}{{{pi ^2}}}left. {left( { – cos frac{{pi t}}{6}} right)} right|_2^3 }={ frac{{36}}{{{pi ^2}}}left( { – cos frac{pi }{2} + cos frac{pi }{3}} right) }={ frac{{36}}{{{pi ^2}}}left( {0 + frac{1}{2}} right) }={ frac{{18}}{{{pi ^2}}} approx 1.82,text{m}}]

Solution.

First we compute the velocity of the object at time (t.) We use the formula

[vleft( t right) = {v_0} + intlimits_0^t {aleft( u right)du} ,]

where (u) is a variable of integration. Hence,

[{vleft( t right) = {v_0} + intlimits_0^t {1 + cos left( {pi u} right)du} }={ {v_0} + left. {left[ {u + frac{{sin left( {pi u} right)}}{pi }} right]} right|_{u = 0}^{u = t} }={ 0 + left( {t + frac{{sin left( {pi t} right)}}{pi }} right) }-{ left( {0 + frac{{sin 0}}{pi }} right) }={ t + frac{{sin left( {pi t} right)}}{pi }.}] Minecraft pe mobile download.

Notice that the velocity (vleft( t right)) is always positive for (t gt 0.) Therefore, the total distance (d) traveled by the object for (1) sec is given by

[{d = intlimits_0^1 {left| {vleft( u right)} right|du} }={ intlimits_0^1 {vleft( u right)du} }={ intlimits_0^1 {left( {u + frac{{sin left( {pi u} right)}}{pi }} right)du} }={ left. {left[ {frac{{{u^2}}}{2} – frac{{cos left( {pi u} right)}}{{{pi ^2}}}} right]} right|_0^1 }={ left( {frac{1}{2} – frac{{cos pi }}{{{pi ^2}}}} right) – left( {0 – frac{{cos 0}}{{{pi ^2}}}} right) }={ frac{1}{2} + frac{2}{{{pi ^2}}} }approx{ 0.70,left( text{m} right).}]

The distance (d) is measured in meters, if the acceleration (a) is measured in meters per second squared.

Solution.

Let the acceleration of the particle be equal to (a,left( {large{frac{text{m}}{{{text{s}^2}}}}normalsize} right)). For motion from rest at constant acceleration, the velocity is given by the formula

Canvas draw 6 0 24. [vleft( t right) = at.]

The distance ({d_1}) traveled by the particle for the (1text{st}) second is written in the form

[{{d_1} = intlimits_0^1 {vleft( t right)dt} }={ intlimits_0^1 {atdt} }={ left. {frac{{a{t^2}}}{2}} right|_0^1 }={ frac{a}{2}.}]

Similarly, we can write down the distance ({d_2}) covered in the (2text{nd}) second:

[{{d_2} = intlimits_1^2 {vleft( t right)dt} }={ intlimits_1^2 {atdt} }={ left. {frac{{a{t^2}}}{2}} right|_1^2 }={ 4a – frac{a}{2} }={ frac{{7a}}{2}.}]

Comparing both expressions, we find that

[{d_2} = 7{d_1},]

where ({d_1}) and ({d_2}) are measured in meters.

Solution.

Integrating the acceleration function gives us the velocity function:

[{vleft( t right) = int {aleft( t right)dt} }={ int {left( {{e^t} – 1} right)dt} }={ {e^t} – t + C.}]

Since ({v_0} = 0,) we have

[{vleft( 0 right) = 0,};; Rightarrow {{e^0} – 0 + C = 0,};; Rightarrow {C = – 1.}]

Hence, the velocity function has the form:

[vleft( t right) = {e^t} – t – 1.]

The average speed of an object is defined as the total distance traveled by the object divided by total time. So, we can write

[{bar v = frac{1}{{{t_2} – {t_1}}}intlimits_{{t_1}}^{{t_2}} {left| {vleft( t right)} right|dt} }={ frac{1}{{{t_2} – {t_1}}}intlimits_{{t_1}}^{{t_2}} {left| {{e^t} – t – 1} right|dt} }={ frac{1}{{2 – 0}}intlimits_0^2 {left( {{e^t} – t – 1} right)dt} }={ frac{1}{2}left. {left( {{e^t} – frac{{{t^2}}}{2} – t} right)} right|_0^2 }={ frac{{{e^2} – 5}}{2} approx 1.19,left( {frac{text{m}}{text{s}}} right)}]

Solution.

We calculate the velocity function from the acceleration function:

[{vleft( t right) = int {aleft( t right)dt} }={ int {left( {5 – 2t} right)dt} }={ 5t – {t^2} + C.}]

To find (C,) we use the initial condition ({v_0} = 6,left( {large{frac{text{m}}{text{s}}}normalsize} right),) so we get

[vleft( t right) = 5t – {t^2} + 6.]

Notice that the velocity is equal to zero when (t =6) seconds. Hence, when calculating the distance, we split the interval of integration into two intervals where the velocity has a constant sign.

Thus, the total distance traveled by the particle is given by

[{d = intlimits_0^8 {left| {vleft( t right)} right|dt} }={ intlimits_0^8 {left| {5t – {t^2} + 6} right|dt} }={ intlimits_0^6 {left( {5t – {t^2} + 6} right)dt} }-{ intlimits_6^8 {left( {5t – {t^2} + 6} right)dt} }={ left. {left( {frac{{5{t^2}}}{2} – frac{{{t^3}}}{3} + 6t} right)} right|_0^6 }-{ left. {left( {frac{{5{t^2}}}{2} – frac{{{t^3}}}{3} + 6t} right)} right|_6^8 }={ 2 cdot 54 – frac{{112}}{3} }={ frac{{212}}{3} }approx{ 70.67,text{m}}]

Solution.

Determine the velocity ({v_1}) at the moment (t = {T_1}:)

[{{v_1} = {a_1}t = {a_1}{T_1} }={ 5 cdot 5 }={ 25,left( {frac{text{m}}{text{s}}} right)}]

On the second stage, the particle's velocity varies according to the equation

[v = {v_1} + {a_2}t,]

where (0 le t le {T_2}.) Since (vleft( {{T_2}} right) = 0,) we can easily find the value of acceleration ({a_2}:)

[{{a_2} = – frac{{{v_1}}}{{{T_2}}} = – frac{{{a_1}{T_1}}}{{{T_2}}} }={ – frac{{25}}{{10}} }={ – 2.5,left( {frac{m}{{{s^2}}}} right)}]

The value of ({a_2}) is negative as the particle is decelerating.

Let's now move on to the calculation of distances. On the first stage, the distance traveled by the particle is equal to

[{{d_1} = intlimits_0^{{T_1}} {{a_1}tdt} }={ {a_1}intlimits_0^{{T_1}} {tdt} }={ left. {frac{{{a_1}{t^2}}}{2}} right|_0^5 }={ frac{{5 cdot {5^2}}}{2} }={ 62.5,text{m}}]

The distance traveled on the second stage is given by

[{{d_2} = intlimits_0^{{T_2}} {left( {{v_1} + {a_2}t} right)dt} }={ intlimits_0^{{T_2}} {left( {{v_1} + {a_2}t} right)dt} }={ left. {left( {{v_1}t + frac{{{a_2}{t^2}}}{2}} right)} right|_0^{10} }={ 25 cdot 10 – frac{{2,5 cdot {{10}^2}}}{2} }={ 125,text{m}}]

Hence, the total distance covered by the particle is

[{d = {d_1} + {d_2} }={ 62.5 + 125 }={ 187.5,text{m}}]

Solution.

Suppose that the ball is thrown upward with an initial velocity of ({v_0}).

The velocity of the ball varies by the law

[vleft( t right) = {v_0} – gt,]

where (g) is the acceleration due to gravity: (g approx 9.8,left( {large{frac{text{m}}{{{text{s}^2}}}}normalsize} right))

The distance traveled in the (1text{st}) second is

[{{d_1} = intlimits_0^1 {left( {{v_0} – gt} right)dt} }={ left. {left( {{v_0}t – frac{{g{t^2}}}{2}} right)} right|_0^1 }={ {v_0} – frac{g}{2}.}]

Respectively, the distance covered for the (2text{nd}) second is given by

[{{d_2} = intlimits_1^2 {left( {{v_0} – gt} right)dt} }={ left. {left( {{v_0}t – frac{{g{t^2}}}{2}} right)} right|_1^2 }={ left( {2{v_0} – 2g} right) – left( {{v_0} – frac{g}{2}} right) }={ {v_0} – frac{{3g}}{2}.}]

We can find ({v_0}) from the condition (large{frac{{{d_1}}}{{{d_2}}}}normalsize = 2.) So

[{frac{{{v_0} – frac{g}{2}}}{{{v_0} – frac{{3g}}{2}}} = 2,};; Rightarrow {{v_0} – frac{g}{2} = 2left( {{v_0} – frac{{3g}}{2}} right),};; Rightarrow {{v_0} – frac{g}{2} = 2{v_0} – 3g,};; Rightarrow {{v_0} = frac{{5g}}{2}.}]

The maximum height is determined by the equation

[H = frac{{v_0^2}}{{2g}}.]

Hence,

[require{cancel}{H = frac{{{{left( {frac{{5g}}{2}} right)}^2}}}{{2g}} }={ frac{{25{g^cancel{2}}}}{{8cancel{g}}} }={ frac{{25g}}{8} }approx{ 30.7,left( text{m} right)}]

Basic Relationships

Recall from our study of derivatives that if (xleft( t right)) is the position of an object moving along a straight line at time (t,) then the velocity of the object is

[vleft( t right) = frac{{dx}}{{dt}},]

and the acceleration is given by

[aleft( t right) = frac{{dv}}{{dt}} = frac{{{d^2}x}}{{d{t^2}}}.]

Using the integral calculus, we can calculate the velocity function from the acceleration function, and the position function from the velocity function.

The Fundamental Theorem of Calculus says that

[{intlimits_{{t_1}}^{{t_2}} {aleft( t right)dt} = left. {vleft( t right)} right|_{{t_1}}^{{t_2}} }={ vleft( {{t_2}} right) – vleft( {{t_1}} right).}]

Similarly, the difference between the position at time ({{t_1}}) and the position at time ({{t_2}}) is determined by the equation

[{intlimits_{{t_1}}^{{t_2}} {vleft( t right)dt} = left. {xleft( t right)} right|_{{t_1}}^{{t_2}} }={ xleft( {{t_2}} right) – xleft( {{t_1}} right).}]

If the object moves from the position (xleft( {{t_1}} right)) to the position (xleft( {{t_2}} right),) the change (xleft( {{t_2}} right) – xleft( {{t_1}} right)) is called the displacement of the object:

[Delta x = xleft( {{t_2}} right) – xleft( {{t_1}} right).]

To find the total distance traveled by the object between time ({{t_1}}) and time ({{t_2}},) we need to compute the integral of (left| {vleft( t right)} right|:)

[d = intlimits_{{t_1}}^{{t_2}} {left| {vleft( t right)} right|dt} .]

Constant Acceleration

Suppose that an object is moving along a straight line with the constant acceleration (a.) At time ({t_1} = 0,) the object has an initial velocity ({v_0}) and an initial position ({x_0}.)

The velocity and position of the object at time (t) are given by the equations

[vleft( t right) = {v_0}t + at,]

[xleft( t right) = {x_0} + {v_0}t + frac{{a{t^2}}}{2}.]

Solved Problems

Click or tap a problem to see the solution.

Distance Using Acceleration And Time

Solution.

Given that the velocity is positive for (t gt 0,) the total distance traveled for the time interval (left[ {0,t} right)) is expressed by the integral:

[{dleft( t right) = intlimits_0^t {left| {vleft( u right)} right|du} }={ intlimits_0^t {vleft( u right)du} }={ intlimits_0^t {sqrt {4 +u} du} ,}]

where (u) is the inner variable which has no impact on the computation of the integral.

Integration yields:

[{dleft( t right) = left. {frac{{2sqrt {{{left( {4 + u} right)}^3}} }}{3}} right|_0^t }={ frac{2}{3}left[ {sqrt {{{left( {4 + t} right)}^3}} – 8} right].}]

Substituting (t = 5,) we have

[{dleft( {t = 5,text{s}} right) = frac{2}{3}left[ {sqrt {{{left( {4 + 5} right)}^3}} – 8} right] }={ frac{2}{3} cdot 19 }={ frac{{38}}{3} approx 12.7,text{m}}]

Solution.

The equation of motion of the particle has the form

[v = frac{{dx}}{{dt}} = 2sqrt x .]

We have a simple differential equation that describes the particle's position as a function of time. Separating the variables and integrating both sides yields:

[{frac{{dx}}{{sqrt x }} = 2dt,};; Rightarrow {int {frac{{dx}}{{sqrt x }}} = 2int {dt} ,};; Rightarrow {2sqrt x = 2t + C,};; Rightarrow {sqrt x = t + C.}]

It follows from the initial condition (xleft( {t = 0} right) = 0) that the (C = 0.) Hence, the particle moves according to the law:

[x = {t^2}.]

It is easy to see that (t = 10,text{s}) when (x = 100,text{m}.)

Solution.

Let's first compute the displacement (Delta x) of the object when (t = 5,text{s}.) Integrating the velocity expression, we obtain

[{Delta x = intlimits_{{t_1}}^{{t_2}} {vleft( t right)dt} }={ intlimits_0^5 {left( {6 – 2t} right)dt} }={ left. {left( {6t – {t^2}} right)} right|_0^5 }={ 5,text{m}.}]

Notice that the velocity changes sign at ({t_1} = 3.) Therefore, to calculate the distance (d) traveled by the object we split the initial interval (left[ {0,5} right]) into two subintervals (left[ {0,3} right]) and (left[ {3,5} right].) This yields

[{d = intlimits_{{t_1}}^{{t_2}} {left| {vleft( t right)} right|dt} }={ intlimits_0^5 {left| {6 – 2t} right|dt} }={ intlimits_0^3 {left| {6 – 2t} right|dt} + intlimits_3^5 {left| {6 – 2t} right|dt} }={ intlimits_0^3 {left( {6 – 2t} right)dt} – intlimits_3^5 {left( {6 – 2t} right)dt} }={ left. {left( {6t – {t^2}} right)} right|_0^3 – left. {left( {6t – {t^2}} right)} right|_3^5 }={ left[ {left( {18 – 9} right) – 0} right] }-{ left[ {left( {30 – 25} right) – left( {18 – 9} right)} right] }={ 9 + 4 }={ 13,text{m}.}]

Solution.

Given that the initial velocity is zero: ({v_0} = 0,) we determine the velocity equation:

[{vleft( t right) = int {aleft( t right)dt} = int {cos frac{{pi t}}{6}dt} }={ frac{6}{pi }sin frac{{pi t}}{6}.}]

The distance traveled in the (3text{rd}) second is

[{d = intlimits_2^3 {vleft( t right)dt} }={ intlimits_2^3 {frac{6}{pi }sin frac{{pi t}}{6}dt} }={ frac{{{6^2}}}{{{pi ^2}}}left. {left( { – cos frac{{pi t}}{6}} right)} right|_2^3 }={ frac{{36}}{{{pi ^2}}}left( { – cos frac{pi }{2} + cos frac{pi }{3}} right) }={ frac{{36}}{{{pi ^2}}}left( {0 + frac{1}{2}} right) }={ frac{{18}}{{{pi ^2}}} approx 1.82,text{m}}]

Finding Distance With Acceleration And Time

Solution.

First we compute the velocity of the object at time (t.) We use the formula

[vleft( t right) = {v_0} + intlimits_0^t {aleft( u right)du} ,]

where (u) is a variable of integration. Hence,

[{vleft( t right) = {v_0} + intlimits_0^t {1 + cos left( {pi u} right)du} }={ {v_0} + left. {left[ {u + frac{{sin left( {pi u} right)}}{pi }} right]} right|_{u = 0}^{u = t} }={ 0 + left( {t + frac{{sin left( {pi t} right)}}{pi }} right) }-{ left( {0 + frac{{sin 0}}{pi }} right) }={ t + frac{{sin left( {pi t} right)}}{pi }.}]

Notice that the velocity (vleft( t right)) is always positive for (t gt 0.) Therefore, the total distance (d) traveled by the object for (1) sec is given by

[{d = intlimits_0^1 {left| {vleft( u right)} right|du} }={ intlimits_0^1 {vleft( u right)du} }={ intlimits_0^1 {left( {u + frac{{sin left( {pi u} right)}}{pi }} right)du} }={ left. {left[ {frac{{{u^2}}}{2} – frac{{cos left( {pi u} right)}}{{{pi ^2}}}} right]} right|_0^1 }={ left( {frac{1}{2} – frac{{cos pi }}{{{pi ^2}}}} right) – left( {0 – frac{{cos 0}}{{{pi ^2}}}} right) }={ frac{1}{2} + frac{2}{{{pi ^2}}} }approx{ 0.70,left( text{m} right).}]

The distance (d) is measured in meters, if the acceleration (a) is measured in meters per second squared.

Solution.

Let the acceleration of the particle be equal to (a,left( {large{frac{text{m}}{{{text{s}^2}}}}normalsize} right)). For motion from rest at constant acceleration, the velocity is given by the formula

[vleft( t right) = at.]

The distance ({d_1}) traveled by the particle for the (1text{st}) second is written in the form

[{{d_1} = intlimits_0^1 {vleft( t right)dt} }={ intlimits_0^1 {atdt} }={ left. {frac{{a{t^2}}}{2}} right|_0^1 }={ frac{a}{2}.}]

Similarly, we can write down the distance ({d_2}) covered in the (2text{nd}) second:

[{{d_2} = intlimits_1^2 {vleft( t right)dt} }={ intlimits_1^2 {atdt} }={ left. {frac{{a{t^2}}}{2}} right|_1^2 }={ 4a – frac{a}{2} }={ frac{{7a}}{2}.}]

Comparing both expressions, we find that

[{d_2} = 7{d_1},]

where ({d_1}) and ({d_2}) are measured in meters.

Solution.

Integrating the acceleration function gives us the velocity function:

[{vleft( t right) = int {aleft( t right)dt} }={ int {left( {{e^t} – 1} right)dt} }={ {e^t} – t + C.}]

Since ({v_0} = 0,) we have

[{vleft( 0 right) = 0,};; Rightarrow {{e^0} – 0 + C = 0,};; Rightarrow {C = – 1.}]

Hence, the velocity function has the form:

[vleft( t right) = {e^t} – t – 1.]

The average speed of an object is defined as the total distance traveled by the object divided by total time. So, we can write

[{bar v = frac{1}{{{t_2} – {t_1}}}intlimits_{{t_1}}^{{t_2}} {left| {vleft( t right)} right|dt} }={ frac{1}{{{t_2} – {t_1}}}intlimits_{{t_1}}^{{t_2}} {left| {{e^t} – t – 1} right|dt} }={ frac{1}{{2 – 0}}intlimits_0^2 {left( {{e^t} – t – 1} right)dt} }={ frac{1}{2}left. {left( {{e^t} – frac{{{t^2}}}{2} – t} right)} right|_0^2 }={ frac{{{e^2} – 5}}{2} approx 1.19,left( {frac{text{m}}{text{s}}} right)}]

Solution.

We calculate the velocity function from the acceleration function:

[{vleft( t right) = int {aleft( t right)dt} }={ int {left( {5 – 2t} right)dt} }={ 5t – {t^2} + C.}]

To find (C,) we use the initial condition ({v_0} = 6,left( {large{frac{text{m}}{text{s}}}normalsize} right),) so we get

[vleft( t right) = 5t – {t^2} + 6.]

Notice that the velocity is equal to zero when (t =6) seconds. Hence, when calculating the distance, we split the interval of integration into two intervals where the velocity has a constant sign.

Thus, the total distance traveled by the particle is given by

[{d = intlimits_0^8 {left| {vleft( t right)} right|dt} }={ intlimits_0^8 {left| {5t – {t^2} + 6} right|dt} }={ intlimits_0^6 {left( {5t – {t^2} + 6} right)dt} }-{ intlimits_6^8 {left( {5t – {t^2} + 6} right)dt} }={ left. {left( {frac{{5{t^2}}}{2} – frac{{{t^3}}}{3} + 6t} right)} right|_0^6 }-{ left. {left( {frac{{5{t^2}}}{2} – frac{{{t^3}}}{3} + 6t} right)} right|_6^8 }={ 2 cdot 54 – frac{{112}}{3} }={ frac{{212}}{3} }approx{ 70.67,text{m}}]

Solution.

Determine the velocity ({v_1}) at the moment (t = {T_1}:)

[{{v_1} = {a_1}t = {a_1}{T_1} }={ 5 cdot 5 }={ 25,left( {frac{text{m}}{text{s}}} right)}]

On the second stage, the particle's velocity varies according to the equation

[v = {v_1} + {a_2}t,]

where (0 le t le {T_2}.) Since (vleft( {{T_2}} right) = 0,) we can easily find the value of acceleration ({a_2}:)

[{{a_2} = – frac{{{v_1}}}{{{T_2}}} = – frac{{{a_1}{T_1}}}{{{T_2}}} }={ – frac{{25}}{{10}} }={ – 2.5,left( {frac{m}{{{s^2}}}} right)}]

The value of ({a_2}) is negative as the particle is decelerating.

Let's now move on to the calculation of distances. On the first stage, the distance traveled by the particle is equal to

[{{d_1} = intlimits_0^{{T_1}} {{a_1}tdt} }={ {a_1}intlimits_0^{{T_1}} {tdt} }={ left. {frac{{{a_1}{t^2}}}{2}} right|_0^5 }={ frac{{5 cdot {5^2}}}{2} }={ 62.5,text{m}}]

The distance traveled on the second stage is given by

[{{d_2} = intlimits_0^{{T_2}} {left( {{v_1} + {a_2}t} right)dt} }={ intlimits_0^{{T_2}} {left( {{v_1} + {a_2}t} right)dt} }={ left. {left( {{v_1}t + frac{{{a_2}{t^2}}}{2}} right)} right|_0^{10} }={ 25 cdot 10 – frac{{2,5 cdot {{10}^2}}}{2} }={ 125,text{m}}]

Hence, the total distance covered by the particle is

[{d = {d_1} + {d_2} }={ 62.5 + 125 }={ 187.5,text{m}}]

Solution.

Suppose that the ball is thrown upward with an initial velocity of ({v_0}).

Given Distance And Time Find Acceleration

The velocity of the ball varies by the law

[vleft( t right) = {v_0} – gt,]

where (g) is the acceleration due to gravity: (g approx 9.8,left( {large{frac{text{m}}{{{text{s}^2}}}}normalsize} right))

The distance traveled in the (1text{st}) second is

[{{d_1} = intlimits_0^1 {left( {{v_0} – gt} right)dt} }={ left. {left( {{v_0}t – frac{{g{t^2}}}{2}} right)} right|_0^1 }={ {v_0} – frac{g}{2}.}]

Respectively, the distance covered for the (2text{nd}) second is given by

[{{d_2} = intlimits_1^2 {left( {{v_0} – gt} right)dt} }={ left. {left( {{v_0}t – frac{{g{t^2}}}{2}} right)} right|_1^2 }={ left( {2{v_0} – 2g} right) – left( {{v_0} – frac{g}{2}} right) }={ {v_0} – frac{{3g}}{2}.}]

We can find ({v_0}) from the condition (large{frac{{{d_1}}}{{{d_2}}}}normalsize = 2.) So

[{frac{{{v_0} – frac{g}{2}}}{{{v_0} – frac{{3g}}{2}}} = 2,};; Rightarrow {{v_0} – frac{g}{2} = 2left( {{v_0} – frac{{3g}}{2}} right),};; Rightarrow {{v_0} – frac{g}{2} = 2{v_0} – 3g,};; Rightarrow {{v_0} = frac{{5g}}{2}.}]

The maximum height is determined by the equation

[H = frac{{v_0^2}}{{2g}}.]

Hence,

The velocity and position of the object at time (t) are given by the equations

[vleft( t right) = {v_0}t + at,]

[xleft( t right) = {x_0} + {v_0}t + frac{{a{t^2}}}{2}.]

Solved Problems

Click or tap a problem to see the solution.

Distance Using Acceleration And Time

Solution.

Given that the velocity is positive for (t gt 0,) the total distance traveled for the time interval (left[ {0,t} right)) is expressed by the integral:

[{dleft( t right) = intlimits_0^t {left| {vleft( u right)} right|du} }={ intlimits_0^t {vleft( u right)du} }={ intlimits_0^t {sqrt {4 +u} du} ,}]

where (u) is the inner variable which has no impact on the computation of the integral.

Integration yields:

[{dleft( t right) = left. {frac{{2sqrt {{{left( {4 + u} right)}^3}} }}{3}} right|_0^t }={ frac{2}{3}left[ {sqrt {{{left( {4 + t} right)}^3}} – 8} right].}]

Substituting (t = 5,) we have

[{dleft( {t = 5,text{s}} right) = frac{2}{3}left[ {sqrt {{{left( {4 + 5} right)}^3}} – 8} right] }={ frac{2}{3} cdot 19 }={ frac{{38}}{3} approx 12.7,text{m}}]

Solution.

The equation of motion of the particle has the form

[v = frac{{dx}}{{dt}} = 2sqrt x .]

We have a simple differential equation that describes the particle's position as a function of time. Separating the variables and integrating both sides yields:

[{frac{{dx}}{{sqrt x }} = 2dt,};; Rightarrow {int {frac{{dx}}{{sqrt x }}} = 2int {dt} ,};; Rightarrow {2sqrt x = 2t + C,};; Rightarrow {sqrt x = t + C.}]

It follows from the initial condition (xleft( {t = 0} right) = 0) that the (C = 0.) Hence, the particle moves according to the law:

[x = {t^2}.]

It is easy to see that (t = 10,text{s}) when (x = 100,text{m}.)

Solution.

Let's first compute the displacement (Delta x) of the object when (t = 5,text{s}.) Integrating the velocity expression, we obtain

[{Delta x = intlimits_{{t_1}}^{{t_2}} {vleft( t right)dt} }={ intlimits_0^5 {left( {6 – 2t} right)dt} }={ left. {left( {6t – {t^2}} right)} right|_0^5 }={ 5,text{m}.}]

Notice that the velocity changes sign at ({t_1} = 3.) Therefore, to calculate the distance (d) traveled by the object we split the initial interval (left[ {0,5} right]) into two subintervals (left[ {0,3} right]) and (left[ {3,5} right].) This yields

[{d = intlimits_{{t_1}}^{{t_2}} {left| {vleft( t right)} right|dt} }={ intlimits_0^5 {left| {6 – 2t} right|dt} }={ intlimits_0^3 {left| {6 – 2t} right|dt} + intlimits_3^5 {left| {6 – 2t} right|dt} }={ intlimits_0^3 {left( {6 – 2t} right)dt} – intlimits_3^5 {left( {6 – 2t} right)dt} }={ left. {left( {6t – {t^2}} right)} right|_0^3 – left. {left( {6t – {t^2}} right)} right|_3^5 }={ left[ {left( {18 – 9} right) – 0} right] }-{ left[ {left( {30 – 25} right) – left( {18 – 9} right)} right] }={ 9 + 4 }={ 13,text{m}.}]

Solution.

Given that the initial velocity is zero: ({v_0} = 0,) we determine the velocity equation:

[{vleft( t right) = int {aleft( t right)dt} = int {cos frac{{pi t}}{6}dt} }={ frac{6}{pi }sin frac{{pi t}}{6}.}]

The distance traveled in the (3text{rd}) second is

[{d = intlimits_2^3 {vleft( t right)dt} }={ intlimits_2^3 {frac{6}{pi }sin frac{{pi t}}{6}dt} }={ frac{{{6^2}}}{{{pi ^2}}}left. {left( { – cos frac{{pi t}}{6}} right)} right|_2^3 }={ frac{{36}}{{{pi ^2}}}left( { – cos frac{pi }{2} + cos frac{pi }{3}} right) }={ frac{{36}}{{{pi ^2}}}left( {0 + frac{1}{2}} right) }={ frac{{18}}{{{pi ^2}}} approx 1.82,text{m}}]

Finding Distance With Acceleration And Time

Solution.

First we compute the velocity of the object at time (t.) We use the formula

[vleft( t right) = {v_0} + intlimits_0^t {aleft( u right)du} ,]

where (u) is a variable of integration. Hence,

[{vleft( t right) = {v_0} + intlimits_0^t {1 + cos left( {pi u} right)du} }={ {v_0} + left. {left[ {u + frac{{sin left( {pi u} right)}}{pi }} right]} right|_{u = 0}^{u = t} }={ 0 + left( {t + frac{{sin left( {pi t} right)}}{pi }} right) }-{ left( {0 + frac{{sin 0}}{pi }} right) }={ t + frac{{sin left( {pi t} right)}}{pi }.}]

Notice that the velocity (vleft( t right)) is always positive for (t gt 0.) Therefore, the total distance (d) traveled by the object for (1) sec is given by

[{d = intlimits_0^1 {left| {vleft( u right)} right|du} }={ intlimits_0^1 {vleft( u right)du} }={ intlimits_0^1 {left( {u + frac{{sin left( {pi u} right)}}{pi }} right)du} }={ left. {left[ {frac{{{u^2}}}{2} – frac{{cos left( {pi u} right)}}{{{pi ^2}}}} right]} right|_0^1 }={ left( {frac{1}{2} – frac{{cos pi }}{{{pi ^2}}}} right) – left( {0 – frac{{cos 0}}{{{pi ^2}}}} right) }={ frac{1}{2} + frac{2}{{{pi ^2}}} }approx{ 0.70,left( text{m} right).}]

The distance (d) is measured in meters, if the acceleration (a) is measured in meters per second squared.

Solution.

Let the acceleration of the particle be equal to (a,left( {large{frac{text{m}}{{{text{s}^2}}}}normalsize} right)). For motion from rest at constant acceleration, the velocity is given by the formula

[vleft( t right) = at.]

The distance ({d_1}) traveled by the particle for the (1text{st}) second is written in the form

[{{d_1} = intlimits_0^1 {vleft( t right)dt} }={ intlimits_0^1 {atdt} }={ left. {frac{{a{t^2}}}{2}} right|_0^1 }={ frac{a}{2}.}]

Similarly, we can write down the distance ({d_2}) covered in the (2text{nd}) second:

[{{d_2} = intlimits_1^2 {vleft( t right)dt} }={ intlimits_1^2 {atdt} }={ left. {frac{{a{t^2}}}{2}} right|_1^2 }={ 4a – frac{a}{2} }={ frac{{7a}}{2}.}]

Comparing both expressions, we find that

[{d_2} = 7{d_1},]

where ({d_1}) and ({d_2}) are measured in meters.

Solution.

Integrating the acceleration function gives us the velocity function:

[{vleft( t right) = int {aleft( t right)dt} }={ int {left( {{e^t} – 1} right)dt} }={ {e^t} – t + C.}]

Since ({v_0} = 0,) we have

[{vleft( 0 right) = 0,};; Rightarrow {{e^0} – 0 + C = 0,};; Rightarrow {C = – 1.}]

Hence, the velocity function has the form:

[vleft( t right) = {e^t} – t – 1.]

The average speed of an object is defined as the total distance traveled by the object divided by total time. So, we can write

[{bar v = frac{1}{{{t_2} – {t_1}}}intlimits_{{t_1}}^{{t_2}} {left| {vleft( t right)} right|dt} }={ frac{1}{{{t_2} – {t_1}}}intlimits_{{t_1}}^{{t_2}} {left| {{e^t} – t – 1} right|dt} }={ frac{1}{{2 – 0}}intlimits_0^2 {left( {{e^t} – t – 1} right)dt} }={ frac{1}{2}left. {left( {{e^t} – frac{{{t^2}}}{2} – t} right)} right|_0^2 }={ frac{{{e^2} – 5}}{2} approx 1.19,left( {frac{text{m}}{text{s}}} right)}]

Solution.

We calculate the velocity function from the acceleration function:

[{vleft( t right) = int {aleft( t right)dt} }={ int {left( {5 – 2t} right)dt} }={ 5t – {t^2} + C.}]

To find (C,) we use the initial condition ({v_0} = 6,left( {large{frac{text{m}}{text{s}}}normalsize} right),) so we get

[vleft( t right) = 5t – {t^2} + 6.]

Notice that the velocity is equal to zero when (t =6) seconds. Hence, when calculating the distance, we split the interval of integration into two intervals where the velocity has a constant sign.

Thus, the total distance traveled by the particle is given by

[{d = intlimits_0^8 {left| {vleft( t right)} right|dt} }={ intlimits_0^8 {left| {5t – {t^2} + 6} right|dt} }={ intlimits_0^6 {left( {5t – {t^2} + 6} right)dt} }-{ intlimits_6^8 {left( {5t – {t^2} + 6} right)dt} }={ left. {left( {frac{{5{t^2}}}{2} – frac{{{t^3}}}{3} + 6t} right)} right|_0^6 }-{ left. {left( {frac{{5{t^2}}}{2} – frac{{{t^3}}}{3} + 6t} right)} right|_6^8 }={ 2 cdot 54 – frac{{112}}{3} }={ frac{{212}}{3} }approx{ 70.67,text{m}}]

Solution.

Determine the velocity ({v_1}) at the moment (t = {T_1}:)

[{{v_1} = {a_1}t = {a_1}{T_1} }={ 5 cdot 5 }={ 25,left( {frac{text{m}}{text{s}}} right)}]

On the second stage, the particle's velocity varies according to the equation

[v = {v_1} + {a_2}t,]

where (0 le t le {T_2}.) Since (vleft( {{T_2}} right) = 0,) we can easily find the value of acceleration ({a_2}:)

[{{a_2} = – frac{{{v_1}}}{{{T_2}}} = – frac{{{a_1}{T_1}}}{{{T_2}}} }={ – frac{{25}}{{10}} }={ – 2.5,left( {frac{m}{{{s^2}}}} right)}]

The value of ({a_2}) is negative as the particle is decelerating.

Let's now move on to the calculation of distances. On the first stage, the distance traveled by the particle is equal to

[{{d_1} = intlimits_0^{{T_1}} {{a_1}tdt} }={ {a_1}intlimits_0^{{T_1}} {tdt} }={ left. {frac{{{a_1}{t^2}}}{2}} right|_0^5 }={ frac{{5 cdot {5^2}}}{2} }={ 62.5,text{m}}]

The distance traveled on the second stage is given by

[{{d_2} = intlimits_0^{{T_2}} {left( {{v_1} + {a_2}t} right)dt} }={ intlimits_0^{{T_2}} {left( {{v_1} + {a_2}t} right)dt} }={ left. {left( {{v_1}t + frac{{{a_2}{t^2}}}{2}} right)} right|_0^{10} }={ 25 cdot 10 – frac{{2,5 cdot {{10}^2}}}{2} }={ 125,text{m}}]

Hence, the total distance covered by the particle is

[{d = {d_1} + {d_2} }={ 62.5 + 125 }={ 187.5,text{m}}]

Solution.

Suppose that the ball is thrown upward with an initial velocity of ({v_0}).

Given Distance And Time Find Acceleration

The velocity of the ball varies by the law

[vleft( t right) = {v_0} – gt,]

where (g) is the acceleration due to gravity: (g approx 9.8,left( {large{frac{text{m}}{{{text{s}^2}}}}normalsize} right))

The distance traveled in the (1text{st}) second is

[{{d_1} = intlimits_0^1 {left( {{v_0} – gt} right)dt} }={ left. {left( {{v_0}t – frac{{g{t^2}}}{2}} right)} right|_0^1 }={ {v_0} – frac{g}{2}.}]

Respectively, the distance covered for the (2text{nd}) second is given by

[{{d_2} = intlimits_1^2 {left( {{v_0} – gt} right)dt} }={ left. {left( {{v_0}t – frac{{g{t^2}}}{2}} right)} right|_1^2 }={ left( {2{v_0} – 2g} right) – left( {{v_0} – frac{g}{2}} right) }={ {v_0} – frac{{3g}}{2}.}]

We can find ({v_0}) from the condition (large{frac{{{d_1}}}{{{d_2}}}}normalsize = 2.) So

[{frac{{{v_0} – frac{g}{2}}}{{{v_0} – frac{{3g}}{2}}} = 2,};; Rightarrow {{v_0} – frac{g}{2} = 2left( {{v_0} – frac{{3g}}{2}} right),};; Rightarrow {{v_0} – frac{g}{2} = 2{v_0} – 3g,};; Rightarrow {{v_0} = frac{{5g}}{2}.}]

The maximum height is determined by the equation

[H = frac{{v_0^2}}{{2g}}.]

Hence,

[require{cancel}{H = frac{{{{left( {frac{{5g}}{2}} right)}^2}}}{{2g}} }={ frac{{25{g^cancel{2}}}}{{8cancel{g}}} }={ frac{{25g}}{8} }approx{ 30.7,left( text{m} right)}]