Question

# If a particle has velocity v(t)=t2âˆ’5t+6, the average speed and the magnitude of average velocity in 6 sec respectively are

A
5518 m/s,3.5 m/s
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B
7118 m/s,4 m/s
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C
5118 m/s,5 m/s
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D
5518 m/s,3 m/s
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Solution

## The correct option is D 5518 m/s,3 m/sGiven, v(t)=t2−5t+6, 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. ⇒ t2−5t+6=0⇒ (t−2)(t−3)=0⇒ t=2,3 sec Also, we know that xf−xi=∫t0v(t)dt=∫t0(t2−5t+6)dt ⇒ xf−xi=t33−5t22+6t Let particle starts from origin i.e xi=0 So, at t=0,xf=0, at t=2,xf=233−5×222+6×2=143 m at t=3,xf=333−5×322+6×3=92 m at t=6,xf=633−5×622+6×6=18 m The motion of the particle can be represented as shown So, the average speed of the particle is 143+(143−92)+(18−92)6=5518 m/s and, average velocity is 186=3 m/s

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