If a particle has velocity v(t)=t2−5t+6, the average speed and the magnitude of average velocity in 6sec respectively are
A
5518m/s,3.5m/s
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B
7118m/s,4m/s
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C
5118m/s,5m/s
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D
5518m/s,3m/s
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Solution
The correct option is D5518m/s,3m/s Given, 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=2,3sec 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=143m at t=3,xf=333−5×322+6×3=92m at t=6,xf=633−5×622+6×6=18m The motion of the particle can be represented as shown
So, the average speed of the particle is 143+(143−92)+(18−92)6=5518m/s and, average velocity is 186=3m/s