A car is speeding while travelling around a circular track of radius 65.0 m. At a particular point in time it has a speed of 40.5 m/s and an angular acceleration of 0.450 rad/s2. What is the total acceleration of the car at that point? 54.4 m/s2 25.2 m/s2 25.6 m/s2 38.6 m/s2
Question
A car is speeding while travelling around a circular track of radius 65.0 m. At a particular point in time it has a speed of 40.5 m/s and an angular acceleration of 0.450 rad/s2. What is the total acceleration of the car at that point? 54.4 m/s2 25.2 m/s2 25.6 m/s2 38.6 m/s2
Solution
The total acceleration of a body moving in a circular path is given by the formula:
a_total = sqrt((a_tangential)^2 + (a_radial)^2)
where:
- a_tangential is the tangential acceleration, which is the product of the radius and the angular acceleration, and
- a_radial is the radial (or centripetal) acceleration, which is the square of the speed divided by the radius.
Given:
- radius (r) = 65.0 m,
- speed (v) = 40.5 m/s, and
- angular acceleration (alpha) = 0.450 rad/s^2,
we can calculate:
a_tangential = r * alpha = 65.0 m * 0.450 rad/s^2 = 29.25 m/s^2,
a_radial = v^2 / r = (40.5 m/s)^2 / 65.0 m = 25.2 m/s^2.
Then, we can substitute these values into the formula for total acceleration:
a_total = sqrt((29.25 m/s^2)^2 + (25.2 m/s^2)^2) = 38.6 m/s^2.
So, the total acceleration of the car at that point is 38.6 m/s^2.
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