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Question

If velocity of a particle is v(t)=(15−3t) m/s, the average speed of the particle in 6 sec is

A
5 m/s
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B
5.5 m/s
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C
6.5 m/s
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D
6 m/s
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Solution

The correct option is C 6.5 m/s
Given, v(t)=153t, we will check if the particle has changed its direction or not within the asked time interval, i.e. the time when velocity of the particle becomes zero.
153t=0 t=5 sec
Also, we know that
xfxi=t0v(t)dt=t0(153t)dt
xfxi=15t3t22
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=5,xf=15×53×522=752=37.5 m
at t=6,xf=15×63×622=36 m
The motion of the particle can be represented as shown

So, the average speed of the particle is 37.5+1.56=396=6.5 m/s

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