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Question

# A particle's position in the x-y plane varies as x(t)=3t3+4t2, y(t)=16t4+2 Find the acceleration vector of the particle after 1 sec of start.

A

18m/s2 ^i+192 m/s2^j

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B

3m/s2 ^i+16 m/s2^j

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C

26m/s2 ^i+192 m/s2^j

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D

7m/s2 ^i+18 m/s2^j

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Solution

## The correct option is C 26m/s2 ^i+192 m/s2^j Particles position vector in x is given as x(t)=3t3+4t2 vx(t)=dx(t)dt=9t2+8t ax(t)=dvx(t)dt=18t+8 At t = 1 ax(t)=18+8=26m/s2 Particles position vector y is given as y(t)=16t4+2 vy=dy(t)dt=64t3 ay=dvy(t)dt=192t2 At t = 1 sec ay=192m/s2 ⇒→a=ax^i+ay^j →a=26m/s2 ^i+192 m/s2^j

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