a body moving with uniform acceleration covers a distance of 14m in first 2 seconds and 88m in next 4 seconds how much distance is travelled by it in the 5th second?

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Solution

Dear student,

The body is moving under uniform acceleration, so the velocity at 2nd and 4th second be v2 and v4 is,

V2= 14 m/s and V4= 88 m/s

Rate of change of velocity=(V4-V2) / Change in time

Uniform acceleration = (88–14)/2

Thus acceleration a=34 ms^-2

Now, distance travelled in 5th second is Veloctiy at t=5 s. is V5,

According the well known formula, v = u + at,

***V5= U2+ atime = 14 m/s +(34(5-2)) = 116 m/s*

*Thus the body will travel 116m in the 5 seconds***.

Regards

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