Question

$\pink{QUESTION}$
a scooter moving at the spped of 12 metre per second is stopped by applying breaks which produces uniform retardation of 0.6 metre per second square.how much distance will be covered by the scooter till it stops
​

Answer 1

Answer:

Using the standard equations of motion we know:

Dist(x)= Initial Velocity x Time + 1/2 Constant Acceleration x Time^2 (Squared)

We know the Initial Velocity (12 .0 m/s) and the Constant Acceleration (-0.6 m/s/s negative for deceleration) but not the time.

But we also know:

Change in Velocity (∆V) = Acceleration x Time (at)

So

∆V = 12 m/s - 0 m/s = a x t = 0.6 m/s^2 x t

Solving for t = (12 m/s)/(0.6 m/s/s) = 20 s

Plugging that into equation 1:

Dist = 12 m/s x 20 s + 1/2 (-0.6m/s/s) x (20 s)^ 2 = 240 m - 120 m = 120 m to stop