Question

A toboggan is traveling down a long a curve which can be approximated by the parabola y=0.01x² . Dtermine the magnitude of its accleration when it reaches point A, where its speed is _a =10 m/s , and it is increasing at the rate of (a_t )_A = 3m/s² .

Answer 1

Concepts and reason

When a particle is moving along a curved path it poses some acceleration. If only the direction of the velocity is changing, it possesses radial acceleration, if the magnitude of velocity is changing it possesses tangential acceleration. If both the direction and magnitude is changing it possesses both tangential and radial acceleration.

The direction of tangential acceleration is along the tangent drawn to the curve at that point or it is parallel to the direction of instantaneous velocity at that point.

The radial acceleration is perpendicular to the tangential acceleration along the radius of curvature the curve.

The radius of curvature of a curve is defined as the radius of an approximately circular arc passing through points on the curve.

Fundamentals

Suppose a particle is moving along a curve $y=f(x) $ with a velocity V.

The expression for tangential acceleration is:

Here, is the derivative of velocity with respect to time.

The expression for radial acceleration is:

Here, is the radius of curvature of the curve.

The resultant acceleration is given as:

The radius of curvature is given as:

Consider the equation of curve given.

$y=0.01x2 $

Calculate the first derivative of above equation, differentiate it with respect to x.

Again, differentiate with respect to x.

Calculate the radius of curvature.

Substitute 0.02 for and for .

…… (1)

Calculate the coordinates of point A.

Substitute for x in equation (1).

Calculate the radial acceleration at point A.

Substitute for and for V.

The tangential acceleration is given as:

Calculate the net acceleration at point A.

$0+,pN=0 $

Substitute for and for .

Ans:

The acceleration of particle when it is at point A is .