Question

A toy car accelerates from rest at $2m{s}^{-2}$ for 3 seconds. Afterwards the car begins to slow down. It travels a combined total of 27 meters before coming to rest. What is its rate of acceleration when stopping and what is the total time of the motion?

• A. $a=-2m{s}^{-2},t=6s$
• B. $a=-1m{s}^{-2},t=9s$
• C. $a=-2m{s}^{-2},t=9s$
• D. $a=-1m{s}^{-2},t=6s$

Answer 1

The correct option is B $a=-1m{s}^{-2},t=9s$

Let the initial velocity of toy car be $0m{s}^{-1}$, velocity to which the toy car is accelerated be $v_{1}m{s}^{-1}$
Given: time of acceleration $t_1=3m{s}^{-1}$
We can divide this question into two parts. One in which the toy car accelerates from $0m{s}^{-1}$ to $v_{1}m{s}^{-1}$
Next, when it decelerates from $v_{1}m{s}^{-1}$ to $0m{s}^{-1}$
Let the total distance travelled be 's' and total time taken be 't'.
Case I: when the car accelerates
Initial velocity, $u_1=0m{s}^{-1}$
acceleration, $a_1=2m{s}^{-2}$ and final velocity be $v_1$
Using $s_1=u_1{t}_1+\frac{1}{2}{a}_1{t}_1^2$
$s_1=0+\frac{1}{2}\times 2\times 3^2=9m$
Now, using $v_1=u_1+a_1{t}_1$,
$v_1=0+2\times 3=6m{s}^{-1}$

Case II: When the car is retarding from $6m{s}^{-1}$ to $0m{s}^{-1}$
Final velocity, $v_2=0$ and initial velocity, $u_1=6m{s}^{-1}$
$a_2$ be the acceleration and $s_2$ be the distance covered.
Using $v_2^2=u_2^2+2a_2{s}_2$
$0^2=6^2+2a_2{s}_2$
$s_2=\frac{-18}{a_2}$
Total distance travelled, $s=27m$
$s=s_1+s_2=9+\frac{-18}{a_2}$ = 27
$\Rightarrow a_2=-1m{s}^{-2}$
We know, $v_2=u_2+a_2{t}_2$
$\therefore 0=6+(-1)t_2$
$\Rightarrow$$t_2=6s$

$\therefore$ Total time:
$t=t_1+t_2$
$t=3+6=9s$