If velocity of a particle is v(t)=t2−5tm/s, the average speed and magnitude of average velocity of the particle in 10sec respectively are
A
252m/s,253m/s
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B
253m/s,252m/s
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C
12512m/s,2512m/s
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D
2512m/s,12512m/s
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Solution
The correct option is A252m/s,253m/s Given, v(t)=t2−5t, 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=0⇒t=0,5sec Also, we know that xf−xi=∫t0v(t)dt=∫t0(t2−5t)dt ⇒xf−xi=t33−5t22 Let particle starts from origin i.e xi=0 So, at t=0,xf=0, at t=5,xf=533−5×522=−1256m at t=10,xf=1033−5×1022=5006m The motion of the particle can be represented as shown
So, the average speed of the particle is |−1256|+|−1256|+500610=252=12.5m/s and, average velocity is 500610=253m/s