Question

an automobie moving at a seed of 72km/h is brought to rest in 2 sec. by application of brakes. How much distance does the automobile covers in these 2 seconds?

Answer 1

initial velocity (v) = 72 km/h
=72×1000/ 60×60
=72000/3600
=20m/s
final velocity (u) /rest = 0
time taken = 2s

therefore, $\green{acceleration \:(a)}$= v - u/ t
= 20 - 0/ 2
= 20/2 = 10m/s
Now, for $\orange{distance\: (s)}$
s= ut+1/2×at²
=>s= 0×2+1/2×10×(2)²
=>s= 0+1×5×4
=>s= 0+20
=>s= 20m