Question

# If velocity of a particle is v(t)=(15−3t) m/s, the average speed of the particle in 6 sec is

A
5 m/s
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B
5.5 m/s
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C
6.5 m/s
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D
6 m/s
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Solution

## The correct option is C 6.5 m/sGiven, v(t)=15−3t, 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. ⇒ 15−3t=0⇒ t=5 sec Also, we know that xf−xi=∫t0v(t)dt=∫t0(15−3t)dt ⇒ xf−xi=15t−3t22 Let particle starts from origin i.e xi=0 So, at t=0,xf=0, at t=5,xf=15×5−3×522=752=37.5 m at t=6,xf=15×6−3×622=36 m The motion of the particle can be represented as shown So, the average speed of the particle is 37.5+1.56=396=6.5 m/s

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