Question

A test car starts from rest on a horizontal circular track of 120-m radius and increases its speed at a uniform rate to reach 110 km/h in 11 seconds. Determine the magnitude a of the total acceleration of the car 9 seconds after the start.

Answer 1

The magnitude of the total acceleration of the car 9 seconds after the start is 5.90 m/s², which is calculated using the vector sum of the tangential acceleration (2.78 m/s²) and the centripetal acceleration (5.21 m/s²).

Step-by-step explanation:

To calculate the total acceleration of the test car on the circular track, we need to determine two components: the tangential acceleration and the centripetal acceleration. The tangential acceleration can be found since the car increases its speed uniformly to reach 110 km/h (which is 30.56 m/s) in 11 seconds.

First, calculate the tangential acceleration (a_t):

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After 9 seconds, the speed of the car (v) will be:

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Now calculate the centripetal acceleration (a_c) using the speed at 9 seconds:

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The magnitude of the total acceleration (a) at 9 seconds is the vector sum of the tangential and centripetal accelerations:

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• a = √(a_t² + a_c²) = √(2.78 m/s²)² + (5.21 m/s²)² = √(7.7284+27.1441) = √34.8725 = 5.90 m/s²

The magnitude of the total acceleration of the car 9 seconds after the start is 5.90 m/s².