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Question

If a particle has velocity v(t)=t25t+6, the average speed and the magnitude of average velocity in 6 sec respectively are

A
5518 m/s,3.5 m/s
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B
7118 m/s,4 m/s
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C
5118 m/s,5 m/s
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D
5518 m/s,3 m/s
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Solution

The correct option is D 5518 m/s,3 m/s
Given, v(t)=t25t+6, 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.
t25t+6=0 (t2)(t3)=0 t=2,3 sec
Also, we know that
xfxi=t0v(t)dt=t0(t25t+6)dt
xfxi=t335t22+6t
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=2,xf=2335×222+6×2=143 m
at t=3,xf=3335×322+6×3=92 m
at t=6,xf=6335×622+6×6=18 m
The motion of the particle can be represented as shown

So, the average speed of the particle is 143+(14392)+(1892)6=5518 m/s
and, average velocity is 186=3 m/s

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