If a particle has velocity v(t)=t2−6t+8, the average speed in 5sec is
A
85m/s
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B
2315m/s
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C
2815m/s
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D
195m/s
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Solution
The correct option is C2815m/s Given, v(t)=t2−6t+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. ⇒t2−6t+8=0⇒(t−2)(t−4)=0⇒t=2,4sec
Also, we know that xf−xi=∫t0v(t)dt=∫t0(t2−6t+8)dt ⇒xf−xi=t33−3t2+8t
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=2,xf=233−3×22+8×2=203m
at t=4,xf=433−3×42+8×4=163m
at t=5,xf=533−3×52+8×5=203m
The motion of the particle can be represented as shown
So, the average speed of the particle is 203+2×(203−163)5=2815m/s