Question

A particle which is moving in a straight line with constant acceleration describes distances of 10 m
and 15 m in two successive seconds. Find the acceleration.
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Answer 1

Answer:

Correct option is

B

50

The distance travelled in nth second is

Sn=u+21(2n−1)a          ....(1)

So distance travelled in tth&(t+1)th second are

St=u+21(2t−1)a          ....(2)

St+1=u+21(2t+1)a      ....(3)

As per question,

St+St+1=100=2(u+at)         ....(4)

Now from first equation of motion the velocity of particle after time t, if it moves with an acceleration a is

v=u+at              ....(5)

where u is initial velocity

So from equation (4) and (5), we get v=50cm/s