Question

on a large slippery ground a boy left his dog sitting and walks away with a constant velocity 2 ms-1 when he is 199m away from dog the dog decides to catch him and there after move together the dog cannot develop acceleration more than a=2 in what minimum time dog will meet the boy


ans=21s

Answer 1

Answer:

Explanation:

Let velocity=v

Initial position u=0

Now

Newton third law

V^2-u^2=2aS

(2)^2-u^2=2×2×199

Hope it helps u now

Answer 2

Given,

v = 2m/s-1

a= 2

u= 0

s= 199m

Now,

You can apply equation no. second and solve this.