Question

A car accelerates from rest at a constant rate of 5 m/${{s}^{2}}$${{s}^{2}}$ for some time after which it decelerates at a constant rate 3 $m/{{s}^{2}}$ to come to rest. If the total time elapsed is 20 s, then the maximum velocity acquired by the car is
A. 5 m/s
B. 37.5 m/s
C. 30.5 m/s
D. 160 m/s

Answer 1

Hint: In this question we have been asked to calculate the maximum velocity acquired by the car during acceleration and deceleration. We have been given the acceleration and deceleration of the car along with the time the car takes to come to rest. Therefore, to solve this question, we use the equation of kinetic motion. First we calculate the velocity during acceleration and next during deceleration. Then we sum up the time required for both the process as it is given as t.

Formula used:
The equation of motion is,
$v= u + at$
Here, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Complete step by step solution:
Given that the car accelerates at 5 $m/{{s}^{2}}$. Now, let the car accelerate for time ${{t}_{1}}$ before decelerating. Therefore, using second equation of motion we get
v = 0 + 5${{t}_{1}}$
On solving ${{t}_{1}}$= $\dfrac{v}{5}$ ……………………………………. (1)
Similarly let ${{t}_{2}}$ be the time it takes to comes to rest after decelerating
Therefore 0 = v - 3${{t}_{2}}$
On solving , we get
${{t}_{2}}$= $\dfrac{v}{3}$………………………………………………… (2)

Now we have been given that the total time taken by the car to come to rest is t
Therefore, t = ${{t}_{1}}$+ ${{t}_{2}}$………………………………………… (3)
From (1),(2) and (3) we get
T = $\dfrac{v}{5}+\dfrac{v}{3}$
On solving, we get
$t=\dfrac{8v}{15}$
$\Rightarrow t=\dfrac{8\times 20}{15}$
$\therefore t= 37.5\,m/s$

Thus, option B is correct.

Note: The equations of motion are a set of equations that are used to describe the motion of a particle or system. These equations relate the parameters such as initial and final velocity, acceleration and time. The equations of motion are applicable to any system having constant acceleration and deceleration.