Question

The initial velocity of particle is 10 m/s and it's retardation is 2 m/s^2. The distance moved by the particle in 4th second of it's motion is what?

Answer 1

Initial velocity = 10 m/s
acceleration = -2 m/s²
time = 4 sec
distance = ?

using second law of uniformly accelerated motion,
we have,
s = ut + 1/2at²
s = 40+ 1/2×-2×16
s = 40 - 16
s = 24 m----(1)

Distance covered till fourth second of motion will be 24 metres.
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taking time = 3 .seconds.

using second equation of uniformly accelerated motion,
we have,
s = ut + 1/2at²
s = 30 + 1/2×-2×9
s = 30 - 9
s = 21 metres.
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Distance covered from 3th second to 4th second.
= 24 - 21
= 3 m