Question

Breaks are applied to a car to produce an acceleration of 6ms- in the opposite direction to the motion. If the car takes 2seconds to stop after the application of breaks. Calculate the distance travelled during this time.

Answer 1

Given
u=6ms
v=0 Because it will come at rest
t=2s
a=?

Frist Find acceleration
a=v-u/t
a=0-6/2
a=-3 which is in form of declaration

s=ut+1/2at^2
s=6×2+1/2×-3×(2)^2
s=12-6
s=6m/s

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Answer 2

a = - 6 m/s^2
t = 2 sec
v = 0
u = ?
 using v = u + at
u = v - at
 u = 0 - (-6 X 2)
u = 12 m/s

speed = Distance/ Time
Distance = Speed X Time
Distance = 12 X 2
                =24 m/s