Question

Brakes are applied to a car to produce a negative acceleration of 6m/s2 .if the car takes 2 s after applying the brakes, calculate the distance it travels during the time.

Answer 1

Acceleration a = -6 m/s^2Time t = 2 sFinal velocity v = 0 m/s
Let initial velocity be uLet distance be s
v = u + atSo, 0 = u + (-6)(2)So, u = 12 m/s
Now, s = ut + (1/2) at^2So, s = 12(2) + (1/2)(-6)(2^2)So, s = 24 - 12So, s = 12 m
Thus, distance travelled is 12 m.
Hope it helps u.

Answer 2

$\sf \pmb{Answer :}$

12 m

Explanation:

Acceleration produced on applying brakes

= -6 m/(s)^2

Time taken to stop = 2 s

Let initial velocity be u m/s

final velocity is 0 m/s

Using equation of motion

v = u + at

0 = u +(-6)×(2)

u = 12 m/s

Now, using

(v)^2 - (u)^2 = 2as

(0)^2 - (12)^2 = 2 (-6)(s)

-144 = -12(s)

s = 12 m