Question

the brakes applied to a car produces an acceleration of 6 m per second square in the opposite direction to the motion if the car takes 2 seconds to stop after the application of the brakes calculate the distance it travels during this time

Answer 1

V = 0 m/s (as brakes have been applied)

a = - 6 m/s² ( as it is opposite to the direction of motion) , t = 2 s

So we need to find d

V = u + at

0 = u + (-6)2 , 0 = u - 12

u = 12 m/s

V² - u² = 2ad

0² - (12)² = 2(-6) d

0 - 144 = - 12d

-144 = -12d, d = -144/-12 = 12m

d = 12m

Answer 2

Here
V is final velocity
U is initial velocity
a is acceleration
T is time
S is distance travelled

If brakes are applied therefore
v = 0
a = - 6m per second square
t = 2 sec

By first equation of motion :-

V = u + at
0 = u + (-6)2
0 + 12 = u
u = 12 m/s

By second equation of motion :-

S = ut + 1/2at*t
S = 12(2) + 1/2(-6)(2)(2)
S = 24 -12
S = 12m