Question

The brakes applied to a car produce a negative acceleration of 20m/s2. If
the car takes 8s to stop after applying the brakes calcute the distance it travels during this time

Answer 1

Given ---

Acceleration (a) = -20 m/s²
Time (t) = 8 sec
Final velocity (v) = 0 m/s.... (because car stops)

To find ---

Initial velocity (u)
Distance covered (s)

We know, by first equation of motion, that

$v = u + at$

$0 = u + ( - 20) \times 8$

$0 = u + ( - 160)$

$- u = - 160$

Cancelling minus sign,

$u = 160$

Hence, car was traveling with a speed of 160 m/s initially.

Now, by third equation of motion,

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

$= > s = \frac{ {v}^{2} - {u}^{2} }{2a}$

$= > s = \frac{ {0}^{2} - {160}^{2} }{2 \times - 20}$

$= > s = \frac{ - 160 \times 160}{ - 40}$

$= > s = \frac{ - 160 \times 4}{ - 1}$

$= > s = 160 \times 4$

$= > s = 640$

$\boxed{Hence,\: the \:car\: travels\: 640 \:m}$

Hope this answer helps you. ;-)