Question

A toy car is powered by water and compressed air. When the car is fully pumped up and released, it accelerates at 2 $m{s}^{-2}$ for 3 seconds. As soon as this water runs out the car begins to slow down. The car travels 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 = - 2 , t = 6 s
• B. a = - 1 , t = 9 s
• C. a = - 2 , t = 9 s
• D. a = - 1 , t = 6 s

Answer 1

The correct option is B

a = - 1 , t = 9 s

Divide this question into two parts. One in which it accelerates from $0m{s}^{-1}$ to $v_{1}m{s}^{-1}$. Next, in which it decelerates from $v_{1}m{s}^{-1}$ to $0m{s}^{-1}$.

1^st Part:

Using second equation of motion, $s=ut+\frac{1}{2}a{t}^2$

$s_1=0+\frac{1}{2}\times 2\times 3^2=9m$

Using first equation of motion, $v=u+at$

$v_1=0+2\times 3=6m{s}^{-1}$

2^nd part:

Now, final velocity $=0$

Using third equation of motion:

$v^2=u^2+2as$

$0^2=6^2+2as$

$s=\frac{-18}{a}$

Now, total distance travelled, $9+\frac{-18}{a}=27$

$a=-1m{s}^{-2}$

And, $0=6+(-1)t_2$

$t_2=6s$

Total time $=3+6=9$ sec