Question

32: A particle moving in a plane as x= 2+ 3t?, y= 3+t3,
then find its acceleration at t= 3sec. (approx.)
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Answer 1

Answer:

18.97m/s×2

Step-by-step explanation:

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Answer 2

The result acceleration at time 3 seconds will be 0 m/$sec^{2}$

Step-by-step explanation:

Given:

X coordinate =  2+3t

y coordinate = 3+ 3t

To find: Acceleration at t = 3sec

Procedure:

The position vector= Xi + Yj

= (2+3t)i + (3+3t)j

Differentiating the above equation with respect to time(t)

dr/dt = d(2+3t)i + (3+3t)j /dt

dr/dt = 3i + 3j

Removing vector: $\sqrt[]{x^{2} +y^{2} }$

Resultant speed: $\sqrt{3^{2}+3^{2} }$ = $\sqrt{18}$ = 3$\sqrt{2}$ m/sec

Now, differentiating the velocity w.r.t time:

ds/dt = dr(3$\sqrt{2}$)/dt

ds/dt = 0

Thus, the resultant acceleration at time 3 seconds will be 0 m/$sec^{2}$

(#SPJ3)