If velocity of a particle is v(t)=(15−3t)m/s, the average speed of the particle in 6sec is
A
5m/s
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B
5.5m/s
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C
6.5m/s
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D
6m/s
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Solution
The correct option is C6.5m/s Given, 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=5sec 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=37.5m at t=6,xf=15×6−3×622=36m The motion of the particle can be represented as shown
So, the average speed of the particle is 37.5+1.56=396=6.5m/s