Question

A particle has an initial velocity of 5.5m/s due east and acceleration of 1m/s2 due west.

The distance covered by it in the sixth second of motion is?

Answer 1

Answer:

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Explanation:

Solution :

At t=5s ,

S=5.5×5−1/2×1×(5)2=15m ,

at t=6s .

S=5.5×6−1/2×1×(6)2=15m ,

at t=5.5s .

S=5.5×5.5−1/2×1×(5.5)2=15.125m ,

Distance covered in 6th second ,

=−2×0.125=0.25m.

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Answer 2

I hope it is helpful

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