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=0⇒vx=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∘