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Question

If a particle has velocity v(t)=t2−6t+8, the average speed in 5 sec is

A
85 m/s
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B
2315 m/s
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C
2815 m/s
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D
195 m/s
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Solution

The correct option is C 2815 m/s
Given, v(t)=t26t+8, 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.
t26t+8=0 (t2)(t4)=0 t=2,4 sec
Also, we know that
xfxi=t0v(t)dt=t0(t26t+8)dt
xfxi=t333t2+8t
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=2,xf=2333×22+8×2=203 m
at t=4,xf=4333×42+8×4=163 m
at t=5,xf=5333×52+8×5=203 m
The motion of the particle can be represented as shown

So, the average speed of the particle is 203+2×(203163)5=2815 m/s

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