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Question

a body moving with a uniform accleration crosses a distance 15m in the second and 23 m second. The displacement in 10 s will be.

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Solution

if 20 m was covered in the 7th second i.e. in the time interval (6,7). Average velocity in the this interval, hence, is 20 m/s. That is also the instantaneous velocity at t = 6.5 s.

Similarly, the instantaneous velocity at t = 8.5 s is 24 m/s.

There's a change in velocity of 4 m/s in 2 s. Therefore, acceleration is 4/2 = 2 m/s2s2.

Middlemost instant of the 15th second is t = 14.5 s. From t = 8.5 s, there are 6 seconds till the required instant and so, there would be a change in velocity of 6*2 = 12 m/s. So, velocity at t = 14.5 s is 24+12 = 36 m/s.

Therefore, distance covered in the 15th second is 36 m.


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