Question

A toy car is powered by water and compressed air. When the car is fully pumped up and released, it accelerates at $2\frac{m}{s^2}$ for 3 seconds. As soon as this water runs out the car begins to slow down. The car 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 = - 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 he accelerates from $0\frac{m}{s}$ to $v_1\frac{m}{s}$. Next, when he decelerates from $v_1\frac{m}{s}$ to 0 m/s.

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

$v_1=0+2\times 3=6\frac{m}{s}$

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, $9+\frac{-18}{a}=27$

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

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

$t_2=6s$

Total time = 3 + 6 = 9 sec