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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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