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TS EAMCET · Physics · Laws of Motion

A particle is moving in xy-plane crosses the origin at time t=0. The equation of motion of the particle is y=4x2. If the velocity of the particle is v=(2i^+2j^) m s-1 and acceleration is a=(aj^) m s-2, then the magnitude of a is

  1. A 8
  2. B 16
  3. C 82
  4. D 32
Verified Solution

Answer & Solution

Correct Answer

(D) 32

Step-by-step Solution

Detailed explanation

Given, Velocity in x-direction is, dxdt=vx=2 m s-1 Velocity in y-direction is, dydt=vy=2 m s-1 Acceleration in x-direction is, d2xdt2=dvxdt=ax=0 Acceleration in y-direction is, d2ydt2=dvydt=ay=a Trajectory relation, y=4x2 On differentiating trajectory…
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