Question

a particle move along the parabolic path x= y^2 + 2y + 2 in such a way that the y component of velocity vector remains 5m/s during the motion. The magnitude of the acceleration of particle is?
1. 50m/s^2
2. 100m/s^2
3. 10√2 m/s^2
4. 0.1m/s^2

Answer 1

Answer on Question #48742, Physics, Mechanics | Kinematics | Dynamics

A particle move along the parabolic path $x = y^2 + 2y + 2$ in such a way that the y component of velocity vector remains $5\mathrm{m/s}$ during the motion. The magnitude of the acceleration of particle is?

1. $50\mathrm{m/s^2}$

2. $100\mathrm{m/s^2}$

3. $10\sqrt{2}\mathrm{m/s^2}$

4. $0.1\mathrm{m/s^2}$

Solution:

$$x = y^2 + 2y + 2$$

The first derivative is

$$\frac{dx}{dy} = 2y + 2$$ $$\frac{dx}{dy} = \frac{dx}{dt} \frac{dt}{dy} = v_x \frac{1}{v_y}$$

Thus,

$$v_x = (2y + 2)v_y$$

Since the y component of velocity remains the same, there is no acceleration along the y component, $a_y = 0$.

The acceleration is

$$a_x = \frac{dv_x}{dt} = \frac{dv_x}{dt} \frac{dy}{dy} = \frac{dv_x}{dy} \frac{dy}{dt}$$ $$\frac{dv_x}{dy} = 2v_y$$

Thus,

$$a_x = 2v_y v_y = 2 \cdot 5 \cdot 5 = 50\ \mathrm{m/s^2}$$

Answer: $1.50\mathrm{m/s^2}$

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