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Question

Starting at time t=0 from the origin with speed 1 ms1, a particle follows a two-dimensional trajectory in the xy 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

A
ax = 1 ms2 implies that when the particle is at the origin, ay=1 ms2
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B
ax=0 implies ay=1 ms2 at all times
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C
at t=0, the particle’s velocity points in the xdirection
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D
ax=0 implies that at t=1 s, the angle between the particle’s velocity and the x axis is 45
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Solution

The correct option is D ax=0 implies that at t=1 s, the angle between the particle’s velocity and the x axis is 45
y=x22
at t=0, x=0,y=0u=1} given
equation of trajectory,
y=x22
On differentiating both sides,
dydt=12.2xdxdt
vy=xvx
Again differentiating wrt time,
ay=dxdt.vx+xax
ay=v2x+xax ....(1)

Option
(A)
If ax=1and particle is at origin i.e (x=0,y=0)
ay=V2x
ay=12=1
At origin, at t=0 the speed =1 m/s is given .

Option (B):
ay=v2x+xax
given that, ax=0
ay=v2x
If ax=0 , vx=constant=1, all the time
ay=12=1 (all the time)

Option (C):
at t=0,x=0
vy=xvx=0
speed=1 m/s
vx=1 m/s

Option (D):
ay=v2x+xax
vy=xvx
ax=0, (given in option 'D')
ay=v2x
If ax=0vx=constant
initially (vx=1 m/s)
ay=12=1
at t=1 s
vy=0+ay×t=1×1=1 m/s
tanθ=vyvx
(θ angle made with x axis)
tanθ=vyvx=11=1
θ=45



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