A body travels 100m in the first 2sec nd 104m in the nxt 4secs.how far will it move in nxt 4sec ?

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EXAMPLE:

QUES: How far will it move in last four seconds?

Details:

A body travels 100m in the first 2sec nd 104m in the nxt 4secs.how far will it move in nxt 4sec ?

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We have Va1 = S1/T1 = 100/2 = 50 mps as the average speed over the first 100 m. Va2 = S2/T2 = 104/4 = 26 mps for the next 104 m. And are asked Va3 = S3/T3 for S3 = Va3*4 = ?.

We can't do this without making some assumptions that might not be warranted.

We assume Aa = constant, a constant deceleration rate throughout. Aa = dV/dT = (Va1 - Va2)/(T2 + T1) = 24/6 = 4 m/s^2 deceleration.

So S3 = Va3 T3 = (V3 + U3)/2 T3 = (U3 - Aa T3 + U3)/2 T3 = U3T3 - 1/2 Aa T3^2 where U3 = V2 - Aa T2 is the initial speed for the last 4 s. And V2 = U2 - Aa T2 and U2 = V1 - Aa T1

So we have embedded initial U and final V speeds. V2 = V1 - Aa T1 - Aa T2; and U3 = V1 - Aa T1 - Aa T2 - Aa T2; so that S3 = ( V1 - Aa T1 - Aa T2 - Aa T2) T3 - 1/2 Aa T3^2 = [V1 - Aa(T1 + 2T2)] T3 - 1/2 Aa T3^2 = [U1 - Aa(2T1 + 2T2)] T3 - 1/2 Aa T3^2 = [100 - Aa(3T1 + 2T2)]T3 - 1/2 Aa T3^2 [Eqn 1]

V1 = U1 - Aa T1. From Va = 50 = (V1 + U1)/2 we have U1 = 100 - Aa T1 So plugging into Eqn 1, we have:

S3 = (100 - 4*(3*2 + 2*4))*4 - (1/2)*4*4^2 = 144 meters. ANS.

I have not checked my work, but the approach is correct. That is, the end speed V marks the initial speed U for the next segment. And the distance S = Va T where Va = (U + V)/2 is the average speed over each segment S.

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