Question

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s² over a 5.0 s interval. what is the tangential component of acceleration for a point on the outer edge of the tire during the 5.0-s interval?

O 0.80 m/s²
O 0.50 m/s²
O 0.30 m/s²
O 33 m/s²

Answer 1

The tangential acceleration of a point on the circumference of the tire, given its radius of 0.40 m and angular acceleration of 2.0 rad/s², is 0.80 m/s².

Step-by-step explanation:

The student is asking about tangential acceleration, which is the linear acceleration of a point on the circumference of a rotating body. Given the radius of a tire (0.40 m) and an angular acceleration (2.0 rad/s²), we can find the tangential acceleration by the formula at = rα, where r is the radius and α is the angular acceleration.

We use the given values and get:
at = 0.40 m × 2.0 rad/s² = 0.80 m/s².

Therefore, the tangential component of acceleration for a point on the outer edge of the tire during the 5.0-s interval is 0.80 m/s².