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Question

An object is moving with constant acceleration. Its velocity is 48m s1 at the end of 10 second and becomes 68m s1 at the end of the 15 second. What would be the distance travelled by the object in 15 second?

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Solution

Apply first kinematic equation of motion.

v=u+at

a=vut2t1=68481510=4ms2

Apply second kinematic equation of motion (from initial velocity, u=0)

v2u2=2as

6820=2×4×s

s=578m

Hence, distance travelled in 15sec is578m


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