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Question

If a particle has velocity v(t)=t2−7t+12, the average speed in 6 sec is

A
506 m/s
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B
5518 m/s
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C
6118 m/s
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D
436 m/s
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Solution

The correct option is B 5518 m/s
Given, v(t)=t27t+12, 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.
t27t+12=0 (t3)(t4)=0 t=3,4 sec
Also, we know that
xfxi=t0v(t)dt=t0(t27t+12)dt
xfxi=t337t22+12t
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=3,xf=3337×322+12×3=272 m
at t=4,xf=4337×422+12×4=403 m
at t=6,xf=6337×622+12×6=18 m
The motion of the particle can be represented as shown

So, the average speed of the particle is 272+(272403)+(18403)6=5518 m/s

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