Question

Sita is driving along a straight highway in her car. At time t = 0, when Sita is moving at 10 m/s in the positive x-direction, she passes a signpost at x = 50 m. Here acceleration is a function of time:
$a = 2m{s^{ - 2}} - \left( {\dfrac{1}{{10}}m{s^{ - 3}}} \right)t$
Where is the car when it reaches the maximum velocity?

Answer 1

Hint: Acceleration is the second derivative of position. Using the initial conditions find the position of the particle. Then, using the differential method find the time at which velocity is maximum and then use the time to find the position.

Complete step-by-step solution -
We have the acceleration of the car as given below.
$a = 2m{s^{ - 2}} - \left( {\dfrac{1}{{10}}m{s^{ - 3}}} \right)t............(1)$
Acceleration is the derivative of velocity, hence, we have:
$\dfrac{{dv}}{{dt}} = 2 - \dfrac{1}{{10}}t$
Multiplying both sides by dt, we have:
$dv = \left( {2 - \dfrac{1}{{10}}t} \right)dt$
Integrating both sides, we have:
$\int {dv} = \int {\left( {2 - \dfrac{1}{{10}}t} \right)dt}$
$v = 2t - \dfrac{1}{{20}}{t^2} + C$
At t = 0, we have the velocity as 10 m/s. Hence, we have:
$10 = 2(0) - \dfrac{1}{{20}}{(0)^2} + C$
$C = 10$
The expression for velocity is as follows:
$v = 2t - \dfrac{1}{{20}}{t^2} + 10...........(2)$
Velocity is the time derivative of position, hence, we have:
$\dfrac{{dx}}{{dt}} = 2t - \dfrac{1}{{20}}{t^2} + 10$
Multiplying both sides by dt, we have:
$dx = \left( {2t - \dfrac{1}{{20}}{t^2} + 10} \right)dt$
Integrating both sides, we have:
$\int {dx} = \int {\left( {2t - \dfrac{1}{{20}}{t^2} + 10} \right)dt}$
$x = {t^2} - \dfrac{1}{{60}}{t^3} + 10t + D$
At t = 0, the position is 50 m, hence, we have:
$50 = {(0)^2} - \dfrac{1}{{60}}{(0)^3} + 10(0) + D$
$D = 50$
Hence, the expression for the position is given as below:
$x = {t^2} - \dfrac{1}{{60}}{t^3} + 10t + 50..............(3)$
To find the maximum velocity, we need to equate the time derivative of velocity, which is acceleration to zero. Hence, we have:
$a = 2 - \dfrac{1}{{10}}t = 0$
Solving for t, we have:
$2 = \dfrac{1}{{10}}t$
Multiplying by 10 on both sides, we have:
$t = 20$ sec
Hence, the value of velocity is maximum at t = 20. Now, let us substitute it in equation (3) to get as follows:
$x = {(20)^2} - \dfrac{1}{{60}}{(20)^3} + 10(20) + 50$
Simplifying, we have:
$x = 400 - \dfrac{{400}}{3} + 200 + 50$
$x = 650 - \dfrac{{400}}{3}$
$x = \dfrac{{1950 - 400}}{3}$
$x = \dfrac{{2350}}{3}$
Hence, the position at which the car has the maximum velocity is $\dfrac{{2350}}{3}$ m.

Note: We proceed with first finding the velocity as acceleration is the derivative of velocity so we use the integration to find the velocity w.r.t time and also velocity is the derivative of position w.r.t time so again integrate the velocity function. for maximum velocity find the time by using the double derivative concept of maxima and minima. Here you might make a mistake by reading the question as to when the car has a maximum velocity, if you just find the time, you won't be given full credits. The question is to find where is the maximum velocity, hence, you need to find the position.