Starting at time t = 0 from the origin with speed 1   m   s - 1 , a particle follows a…

Starting at time t=0 from the origin with speed 1 m s-1, a particle follows a two-dimensional trajectory in the x-y plane so that its coordinates are related by the equation y=x22. The x and y components of its acceleration are denoted by ax and ay, respectively. Then
  1. ax=1 m s-2 implies that when the particle is at the origin, ay=1 m s-2
  2. ax=0 implies ay=1 m s-2 at all times
  3. at t=0, the particle's velocity points in the x-direction
  4. ax=0 implies that at t=1s, the angle between the particle's velocity and the x axis is 45°

Solution

y=x22

dydt=12×2xdxdt

vy=xvx

dvydt=dxdt×dxdt+xd2vxdt2

ay=vx2+xax

If x=0, ay=vx2=1, No matters what is the value of ax hence, Option A is correct.

If ax=0 then x=vxt=t

at t=1; x=1 m

then angle made by velocity vector with x -axis

tanθ=vyvx=x=1⇒θ=45°

^

Asked in: JEE Advanced 2020 (Paper 2)

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