JEE Challenger
More from Motion in a Plane
PhyMedAdv 2020 • Paper 2

Motion of Particle along Parabolic Path in Two Dimensions

At time t=0t = 0, a particle begins moving from the origin with an initial speed of 1 ms−11\text{ ms}^{-1} along a two-dimensional trajectory in the xx-yy plane obeying y=x22y = \frac{x^2}{2}. Let axa_x and aya_y represent the components of acceleration along the xx and yy axes, respectively. Which of the following options is/are correct?

Options

A

ax=1 ms−2a_x = 1\text{ ms}^{-2} implies that ay=1 ms−2a_y = 1\text{ ms}^{-2} when the particle is at the origin

Correct
B

ax=0a_x = 0 implies that ay=1 ms−2a_y = 1\text{ ms}^{-2} for all times

Correct
C

at t=0t = 0, the velocity vector of the particle points along the xx-direction

Correct
D

ax=0a_x = 0 implies that at t=1 st = 1\text{ s}, the angle between the particle's velocity vector and the xx-axis is 45∘45^\circ

Correct

Topics & Concepts

Motion in a PlaneKinematics

Step-by-Step Solution

To determine which options are correct, we analyze the kinematic relations for the particle moving along the parabolic path.

1. Relation Between Coordinates and Velocities

The equation of the trajectory is given by: y=x22y = \frac{x^2}{2}

Differentiating both sides with respect to time tt: vy=dydt=ddt(x22)=xdxdt=xvxv_y = \frac{dy}{dt} = \frac{d}{dt}\left(\frac{x^2}{2}\right) = x \frac{dx}{dt} = x v_x

2. Analysis of Option (C)

At time t=0t = 0, the particle is at the origin, so x=0x = 0. Substituting x=0x = 0 into the expression for vyv_y: vy(0)=0⋅vx(0)=0v_y(0) = 0 \cdot v_x(0) = 0

The magnitude of the initial velocity (speed) is given as 1 m s−11\text{ m s}^{-1}: v(0)=vx(0)2+vy(0)2=vx(0)2+0=∣vx(0)∣=1 m s−1v(0) = \sqrt{v_x(0)^2 + v_y(0)^2} = \sqrt{v_x(0)^2 + 0} = |v_x(0)| = 1\text{ m s}^{-1}

Since vy(0)=0v_y(0) = 0 and ∣vx(0)∣=1 m s−1|v_x(0)| = 1\text{ m s}^{-1}, the velocity vector at t=0t = 0 is directed purely along the xx-axis. Therefore, Option (C) is correct.

3. Relation Between Accelerations

Differentiating the velocity relation vy=xvxv_y = x v_x with respect to time tt: ay=dvydt=ddt(xvx)=(dxdt)vx+x(dvxdt)a_y = \frac{dv_y}{dt} = \frac{d}{dt}(x v_x) = \left(\frac{dx}{dt}\right) v_x + x \left(\frac{dv_x}{dt}\right) ay=vx2+xaxa_y = v_x^2 + x a_x

4. Analysis of Option (A)

When the particle is at the origin, x=0x = 0. Substituting x=0x = 0 into the acceleration relation: ay=vx(0)2+0⋅ax=(1)2=1 m s−2a_y = v_x(0)^2 + 0 \cdot a_x = (1)^2 = 1\text{ m s}^{-2}

Since ay=1 m s−2a_y = 1\text{ m s}^{-2} at the origin regardless of the value of axa_x, the statement "ax=1 m s−2a_x = 1\text{ m s}^{-2} implies that ay=1 m s−2a_y = 1\text{ m s}^{-2} when the particle is at the origin" is mathematically true. Therefore, Option (A) is correct.

5. Analysis of Option (B)

If ax=0a_x = 0, the xx-component of acceleration is zero for all times, which means vxv_x is constant: vx(t)=vx(0)=±1 m s−1  ⟹  vx2=1 m2s−2v_x(t) = v_x(0) = \pm 1\text{ m s}^{-1} \implies v_x^2 = 1\text{ m}^2\text{s}^{-2}

Substituting ax=0a_x = 0 and vx2=1v_x^2 = 1 into the acceleration expression: ay(t)=1+x(0)=1 m s−2a_y(t) = 1 + x(0) = 1\text{ m s}^{-2}

This result holds for all times tt. Therefore, Option (B) is correct.

6. Analysis of Option (D)

Given ax=0a_x = 0, assuming without loss of generality that the particle moves in the positive xx-direction (vx=1 m s−1v_x = 1\text{ m s}^{-1}): x(t)=x(0)+vxt=tx(t) = x(0) + v_x t = t

The yy-component of velocity as a function of time is: vy(t)=x(t)vx(t)=t⋅1=t m s−1v_y(t) = x(t) v_x(t) = t \cdot 1 = t\text{ m s}^{-1}

At t=1 st = 1\text{ s}: vx(1)=1 m s−1v_x(1) = 1\text{ m s}^{-1} vy(1)=1 m s−1v_y(1) = 1\text{ m s}^{-1}

Let θ\theta be the angle between the velocity vector and the xx-axis at t=1 st = 1\text{ s}: tan⁡θ=vy(1)vx(1)=11=1  ⟹  θ=45∘\tan\theta = \frac{v_y(1)}{v_x(1)} = \frac{1}{1} = 1 \implies \theta = 45^\circ

Therefore, Option (D) is correct.

Conclusion

Options A, B, C, and D are all correct.

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