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Question

If velocity of a particle is v(t)=t2−5t m/s, the average speed and magnitude of average velocity of the particle in 10 sec respectively are

A
252 m/s,253 m/s
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B
253 m/s,252 m/s
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C
12512 m/s,2512 m/s
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D
2512 m/s,12512 m/s
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Solution

The correct option is A 252 m/s,253 m/s
Given, v(t)=t25t, 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=0 t=0,5 sec
Also, we know that
xfxi=t0v(t)dt=t0(t25t)dt
xfxi=t335t22
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=5,xf=5335×522=1256 m
at t=10,xf=10335×1022=5006 m
The motion of the particle can be represented as shown

So, the average speed of the particle is |1256|+|1256|+500610=252=12.5 m/s
and, average velocity is 500610=253 m/s

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