The position vector, r , of a particle moving in the x-y plane, is given by the

Question

The position vector, r , of a particle moving in the x-y plane, is given by the following equation: r = ˆ i(t 3 12t + 2) + ˆ j(t 2 4t + 4), where 100 t +100 and t is in seconds.

Find the following:

(i) By differentiating r with respect to t, find expressions for the x and y components of the velocity, v at some general time t.

(ii) At what time(s), t, is the instantaneous velocity zero?

(iii) Calculate the instantaneous acceleration at some general time t, expressing your answer as a vector in Cartesian form. (iv) At what time(s) is the particle moving parallel to the y-axis?

Explanation / Answer

dr/dt=d((t3-12t+2)i+j(t2-4t+4))/dt = (3t2-12)i +( 2t-4)j m/s

X component of velocity = 3t2-12 m/s

Y component of velocity = 2t-4 m/s

The instantaneous velocity (3t2-12)2 + (2t-4)2

Instantanoeos velocity =0

(3t2-12)2 +(2t-4)2 =0

9t4-68t2-16t+160=0

Solving this we get value of t when instantaneous velocity =0

Real solution t=2s

Instantaneous acceleration along x axis

d(3t2-12)/dt=6t

Along y axis

d(2t-4)/dt=2

Instantaneous acceleration 6t i +2j m/s2

When particle moves parallel to y axis

X component becomes constant velocity at x component =0

3t2-12=0

3(t2-4)=0

t=+2or,-2

But time cannot be negative so t=2s

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