Question

# If velocity of a particle is v(t)=t2−3t m/s, the average speed and magnitude of average velocity of the particle in 6 sec respectively are

A
6 m/s,3 m/s
B
4.5 m/s,4.5 m/s
C
4.5 m/s,3 m/s
D
3 m/s,3 m/s

Solution

## The correct option is C 4.5 m/s,3 m/sGiven, v(t)=t2−3t, 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. ⇒ t2−3t=0⇒ t=0,3 sec Also, we know that xf−xi=∫t0v(t)dt=∫t0(t2−3t)dt ⇒ xf−xi=t33−3t22 Let particle starts from origin i.e xi=0 So, at t=0,xf=0,  at t=3,xf=333−3×322=−92 m at t=6,xf=633−3×622=18 m So, the average speed of the particle is |−92|+|−92|+186=276=92=4.5 m/s and, average velocity is 186=3 m/sPhysics

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