Question

An automobile whose speed is increasing at a rate of 0.800 m/s2 travels along a circular road of radius 10.0 m.

(a) When the instantaneous speed of the automobile is 3.00 m/s, find the tangential acceleration component.

(b) What is the centripetal acceleration component?

(c) Determine the magnitude and direction of the total acceleration.

Answer 1

(a)
`a_t = 0.800 m/s^2`

The tangential acceleration component of the car is simply equal to the change of the tangential speed divided by the time taken:

`a_t = (\Delta v)/(\Delta t)`

This rate of change is already given by the problem, 0.800 m/s^2, so the tangential acceleration of the car is

`a_t = 0.800 m/s^2`

(b)
`a_c = 0.9 m/s^2`

The centripetal acceleration component is given by

`a_c = (v^2)/(r)`

where

v is the tangential speed

r is the radius of the trajectory

When the speed is v = 3.00 m/s, the centripetal acceleration is (the radius is r = 10.0 m):

`a_c = ((3.00 m/s)^2)/(10.0 m)=0.9 m/s^2`

(c)
`1.2 m/s^2, 48.4^(\circ)`

The centripetal acceleration and the tangential acceleration are perpendicular to each other, so the magnitude of the total acceleration can be found by using Pythagorean's theorem:

`a=√(a_t^2+a_c^2)=√((0.8 m/s^2)^2+(0.9 m/s^2)^2)=1.2 m/s^2`

and the direction is given by:

`tan \theta =(a_c)/(a_t)=(0.9 m/s^2)/(0.8 m/s^2)=1.125\\\theta=tan^(-1)(1.125)=48.4^(\circ)`

where the angle is measured with respect to the direction of the tangential acceleration.