Question

A toy car accelerates from rest at $2m/s^2$ for 3 seconds. Afterwards, the car begins to slow down. It travels a total of 27 m before coming to rest. Find its acceleration while stopping and 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’s break the motion into two parts:
Part 1 from A to B where the toy car is accelerating
Part 2 from B to C where the toy car is decelerating

For the motion from A from B,

Given:

$(u=0m/s,a=2m{s}^{-2},t=3sec,s=?$

Using the second equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

$s=0\times 3+0.5\times 2\times 3^2$

s = 9 m

Now using the first equation of motion,

$v=u+at$

$v=0+2\times 3=6m/s$

For the motion from B to C,

$u=6m/s,s=27-9=18m,a=?,t=?,v=0m/s$

Using the third equation of motion,

$v^2-u^2=2as$

$0-6^2=2\times a\times 18$

$a=-1m/s^2$

From B from C,

$u=6m/s,s=27-9=18m,a=-1m{s}^{-2},t=?,v=0m/s$

Using the first equation of motion,

v = u + at

$0=6-1\times t$
t = 6 sec

Total time = 3 s + 6 s = 9 s