A classical mechanics problem by Sanjay Balaji

using the equation of motion:s=ut+(1/2)at^2 where s is the distance travel , a =acceleration,u=initial velocity,t=time after putting in the given values we get: 400=0 t+0.5a(20)^2 400=0.5a 400 dividing both sides by 400 we get 1=0.5a 1/0.5=a 2=a therefore the acceleration=2m/s^2

From Newton's laws of motion D = Vot + 0.5 at^2 Since the cars goes form the rest then Vo = Zero 400 = 0.5 X 20^2 a Then a = 400 / 200 = 2 m/s^2

s = at^2

Sir, you have put the dimensions incorrect, it should be $m/s^2$ , and not $m/s$

s = ut + 0.5at^2

s=ut+(a t t)/2 as the body is at rest so u=0 s=(a t t)/2 400m=(a 20sec 20sec)/2 or 800m=400a a=2m/sec^2

use equation of motion

s= 1/2 a t^2 s= 400 m t= 20 s

Then a = 2

acceleration= velocity\time

$s(t)\quad =\quad \frac { 1 }{ 2 } a{ t }^{ 2 }\\ s(20)\quad =\quad 400\quad =\quad \frac { 1 }{ 2 } a{ (20) }^{ 2 }\\ a\quad =\quad 2$

s=ut+1/2at^2 u=0 400=1/2a 400 a=400/400 2=2 a=2m/s^2