Question

A displacement of particle at any instant is given by x=8t2-3t3 . calculate the average velocity in the interval t=0 and t=2s and in the time interval t=0 to t=3s x is in m.

Answer 1

Instantaneous velocity =dx/dt
=d(8t^2-3t^3)/dt= 16t-9t^2.
Therefore, v at t= 0 is 0 by putting t= 0 in the above equation. V= 16(2)-9(2^2)
=32-36
= -4m/s
Average velocity= (V+v)/2
=-4/2 = -2 m/s
Then put t=3 in the equation 16t-9t^2 and calculate the second average velocity

Answer 2

Answer:

Instantaneous velocity =dx/dt

=d(8t^2-3t^3)/dt= 16t-9t^2.

Therefore, v at t= 0 is 0 by putting t= 0 in the above equation. V= 16(2)-9(2^2)

=32-36

= -4m/s

Average velocity= (V+v)/2

=-4/2 = -2 m/s

Then put t=3 in the equation 16t-9t^2 and calculate the second average velocity

Explanation:

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