Question

A particle moves along the parabolic path y=ax^2 in such a way that x component of velocity is constant, then acceleration is

Answer 1

Answer:

Step-by-step explanation:

y =ax^2

velocity is given by:dy/dt = 2axdx/dt as x component of the velocity remains constant i.e. dx/dt = c

Vy = (2ac)x

Acceleration is given by:dVy/dt = 2acdx/dt

ay = 2ac2y =ax2velocity is given by:dydt = 2ax dx/dt

as x component of the velocity remains constant i.e.

dxdt = cVy = (2ac)xAcceleration is given by:dVy/dt = 2acdx/dtay = 2ac2  

Acceleration has only y component, so acceleration of the particle is in y direction

i.e.

a→ = 2ac2 j⌢

Answer 2

Answer:

Acceleration along x-axis is zero

Step-by-step explanation:

Step 1: The Given equation is y=a x^{2}

Step 2: Differentiate on both sides with respect to time.

dy/dt=a(2x) dx/dt  

vy=2axvx

Step 3:   dy/dt=velocity along y axis (vy)

dx/dt=velocity along x-axis(vx)

Step 4: Again differentiate on both sides:

dvy/dt=2a vx.dx/dt   [vx is constant]

Step 5:  ay=2a.vx.vx       [ay is the acceleration along y axis]

$\underline{a} \mathrm{y}=2 \mathrm{a} \times(\mathrm{VX})^{2}$

Step 6: Therefore velocity along x-axis is constant

Acceleration along x-axis is zero

Step 7: Therefore acceleration is only along y-axis is $\mathrm{a}=2 \mathrm{a} \times(\mathrm{VX})^{2}$