Starting at time t=0 from the origin with speed 1 ms−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

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ax = 1 ms−2 implies that when the particle is at the origin, ay=1 ms−2

ax = 0 implies ay=1 ms−2 at all times

at t = 0, the particle’s velocity points in the 𝑥-direction

ax = 0 implies that at t = 1 s, the angle between the particle’s velocity and the x axis is 45°

answer is A.

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Detailed Solution

y=x22 at t=0x=0,y=0u=1 given y=x22dydt=12⋅2xdxdt⇒vy=xvxdifferentiate wrt timeay=dxdt⋅Vx+xaxay=vx2+xaxOption(A) If ax = 1 and particle is at origin(x = 0, y =0)ay=vx2ay = 12 = 1At origin, at t = 0 secspeed = 1 (given)(B) Optionay=vx2+xax given in option B,ax=0⇒ay=vx2 If ax=0,vx= constant =1,( all the time )⇒ay=12=1 (all the time) (C) at t=0,x=0 vy=xvx speed =1vy=0vx=1D) ay=vx2+xaxvy=xvxax=0 (given in D option) ⇒ay=vx2 If ax=0⇒Vx= constant initially vx=1⇒ay=12=1 at t=1secvy=0+ay×t=1×1=1tan⁡θ=vyvx=x(θ→ angle with x axis )tan⁡θ=vyvx=11=1θ=45∘

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