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A body starts from rest and accelerates with acceleration a and distance is S after it moves constant speed in time t. and it deaccelerates with a/2.and total distance is 15SFind S |
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Answer» BODY starts from rest so , initial VELOCITY, u=0, let FINAL velocity be v acceleration= a distance covered= S we know, v^2=u^2+2aS or, v^2=0+2aS or, v= sqrt(2aS) Then it moves with constant SPEED for time t. let distance covered be S1 speed=distance/time so, distance=speed*time S1=sqrt(2aS)*t Then it retards with a/2 till it comes to rest let the distance covered be S2 again applying, v^2=u^2+2aS 0=2aS-aS2 S2= 2S. total distance covered is 15S by the problem, S+S1+S2=15S or, S+ t*sqrt2aS+ 2S= 15S or, t*sqrt2aS= 12S or,t^2*(2aS)=144S^2 or, S=(a*t^2)/72 |
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