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Question

If velocity of a particle is v(t)=(5t−10) m/s, the average speed and magnitude of average velocity of the particle in 4 sec respectively are

A
5 m/s,4 m/s
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B
5 m/s,0 m/s
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C
3 m/s,5 m/s
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D
0 m/s,3 m/s
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Solution

The correct option is B 5 m/s,0 m/s
Given, v(t)=5t10, 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.
5t10=0 t=2 sec
Also, we know that
xfxi=t0v(t)dt=t0(5t10)dt
xfxi=5t2210t
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=2,xf=5×22210×2=10 m
at t=4,xf=5×42210×4=0 m
The motion of the particle can be represented as shown

So, the average speed of the particle is 2×|10|4=204=5 m/s
and, average velocity is 04=0 m/s

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