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Question

A particle moves along the curve y=x22. Here x varies with time as x=t22. Where x and y are measured in meter and t in second. At t=2 s, the velocity of the particle (in ms1) is

A
2^i4^j
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B
2^i+4^j
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C
4^i+2^j
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D
4^i2^j
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Solution

The correct option is B 2^i+4^j
Step 1: Calculate Vx
Velocity in x direction is given by:
Vx=dxdt=ddt(t22) =tm/s

at t=2s, Vx=2m/s

Step 2: Calculate Vy
y=x22=t48
Velocity in y direction is given by:
Vy=dydt=ddt(t48) =t32m/s

at t=2s, Vy=4m/s

V=2^i+4^jm/s

Hence the velocity of the particle at t=2s is V=2^i+4^jm/s. Option B is correct.

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