If velocity of a particle is v(t)=2t−4m/s, the average speed and magnitude of average velocity of the particle in 10sec respectively are
A
4.8m/s,4m/s
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B
6m/s,6m/s
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C
6.8m/s,6m/s
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D
5.2m/s,4.4m/s
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Solution
The correct option is C6.8m/s,6m/s Given, v(t)=2t−4, 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. ⇒2t−4=0⇒t=2sec
Also, we know that xf−xi=∫t0v(t)dt=∫t0(2t−4)dt ⇒xf−xi=t2−4t
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=2,xf=22−4×2=−4m
at t=10,xf=102−4×10=60m
The motion of the particle can be represented as shown
So, the average speed of the particle is |−4|+|−4|+6010=6810=6.8m/s
and, average velocity is 6010=6m/s