Question

In a test of safety restraint, a test car was slowed from $5m/s$ to rest in a time of $0.40seconds$.
(a) What acceleration was this?
(b) How far did the car travel during this time?

Answer 1

(a)

Step 1: Given data

Initial velocity, $u=5m/s.$
Final velocity, $v=0m/s.$
Time, $t=0.40seconds$

Step 2: Calculating acceleration
Acceleration, $a=\frac{v-u}{t}=\frac{0-5}{0.40}=-12.5m/s^2.$
As acceleration is negative it is retardation at the rate of $12.5m/s^2$

(b)

Step 1: Given data

Acceleration, $a=12.5m/s^2$
Time,$t=0.40seconds$
Initial velocity, $u=5m/s.$

Step 2: Calculating the distance travelled by car during this time using the second equation of motion.
Distance travelled by test car, $s=u\times t+\frac{1}{2}\times a\times t^2$
$s=(5\times 0.40)+(\frac{1}{2}\times 12.5\times 0.{40}^2)=2+1=3metre$
Hence, the car travelled $3metre$ in this time.