If velocity of a particle is v(t)=(5t−10)m/s, the average speed and magnitude of average velocity of the particle in 4sec respectively are
A
5m/s,4m/s
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B
5m/s,0m/s
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C
3m/s,5m/s
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D
0m/s,3m/s
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Solution
The correct option is B5m/s,0m/s Given, v(t)=5t−10, 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. ⇒5t−10=0⇒t=2sec
Also, we know that xf−xi=∫t0v(t)dt=∫t0(5t−10)dt ⇒xf−xi=5t22−10t
Let particle starts from origin i.e xi=0
So, at t=0,xf=0,
at t=2,xf=5×222−10×2=−10m
at t=4,xf=5×422−10×4=0m
The motion of the particle can be represented as shown
So, the average speed of the particle is 2×|−10|4=204=5m/s
and, average velocity is 04=0m/s