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Question

If velocity of a particle is v(t)=2t−4 m/s, the average speed and magnitude of average velocity of the particle in 5 sec respectively are

A
145 m/s,95 m/s
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B
135 m/s,95 m/s
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C
135 m/s,1 m/s
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D
145 m/s,1 m/s
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Solution

The correct option is C 135 m/s,1 m/s
Given, v(t)=2t4, 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.
2t4=0 t=2 sec
Also, we know that
xfxi=t0v(t)dt=t0(2t4)dt
xfxi=t24t
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=2,xf=224×2=4 m
at t=5,xf=524×5=5 m
The motion of the particle can be represented as shown

So, the average speed of the particle is |4|+|4|+55=135 m/s
and, average velocity is 55=1 m/s

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