Question

An object has a position given by r=(1.30+4.2t^2)i + (3.5-0.75t^3)j, where quantities are in SI units. What is the magnitude of the acceleration of the object at time=2.5s? Please show steps involved.

Answer 1

Position of object is given by:

r = (1.30 + 4.2t^2) i + (3.5 - 0.75t^3)j

Now Velocity is given by:

V = dr/dt

V = d[(1.30 + 4.2t^2) i + (3.5 - 0.75t^3)j]/dt

V = (0 + 2*4.2t) i + (0 - 3*0.75t^2) j

V = 8.4t i - 2.25t^2 j

Now acceleration is given by:

a = dV/dt

a = d[8.4t i - 2.25t^2 j]/dt

a = 8.4*1 i - 2*2.25*t j

a = 8.4 i - 4.5t j

Now at the time t = 2.5 sec, acceleration will be

a = 8.4 i - 4.5*2.5 j

a = 8.4 i - 11.25 j

Magnitude of acceleration will be

|a| = sqrt (8.4^2 + (-11.25)^2)

|a| = 14.04 m/sec^2

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