Question

A particle starting from rest is at x=0, at t=0, the acceleration varies with displacement as shown -- a=24x^(1/3). Then the avg acceleration of the particle from t=0 to 2 sec

Answer 1

The derivative of displacement with respect to time is velocity, therefore:

dxdt=3t2−8t−5dxdt=3t2−8t−5

v=3t2−8t−5v=3t2−8t−5

The velocity at the beginning can be obtained by plugging t=0t=0 in the above relation, which would give:

v=−5m/sv=−5m/s

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