The box slides down the slope described by the equation y 0

The box slides down the slope described by the equation y = (0.05x^2) m, where x is in meters. If the box has x components of velocity and acceleration of v_x = -3 m/s and a_x = -1.5 m/s^2 at x = 5 m, determine v_y and a_y of the box at this instant.

Solution

As per the given equation of the curve, we have y = 0.05x^2

If we differentiate both sides of the equation above, we would obtain a relation as:

dy/dt = 0.1x dx/dt [Equation 1]

Now, we know that the rate of change of x and y is the velocity of the particle along x and y directions respectively.Using this in equation 1 above, we get:

V_y = 0.1xV_x

For the given instant, we have, Vx = -3 m/s and x = 5 m. Using these values in the relation obtained, we get:

Vy = -0.1 x 5 x 3 = -1.5 m/s

Further, if we differentiate the equation 1 once again, we would obtain:

d²y/dt² = 0.1 x d²x/dt² + 0.1 (dx/dt)²

We know that double differentiation of y and x with respect to time is the acceleration along y and x respectively. Hence,

Ay = 0.1 x Ax + 0.1 Vx²

Using the values given and obtained above, we get:

Ay = -0.1 x 5 x 1.5 + 0.1 x 9 = 0.9 - 0.75 = 0.15 m/s^2