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A car is moving with a constant velocity around a circular path. If the radius of the circular path is 48.2 m and the centripetal acceleration is 8.05 m/s2, what is the tangential speed of the car?

User Zohab Ali
by
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2 Answers

1 vote

Answer:

The tangential speed of the car in this circular path is 19.7 m/s

Step-by-step explanation:

The tangential speed of any object in a circular motion with constant velocity is given by the square root of the centripetal acceleration by the radius of the path it's following. So in order to solve this quetion we can use the following formula:

v = sqrt(R*acp) = sqrt(48.2*8.05) = sqrt(388.01) = 19.7 m/s

The tangential speed of the car in this circular path is 19.7 m/s

User Halcwb
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4.9k points
3 votes

Answer:

19.698 m/s

Step-by-step explanation:

From the question,

The expression for tangential speed is given as,

v = √(ar).................... Equation 1

Where v = tangential speed, a = centripetal acceleration, r = radius of the circular path.

Given: a = 8.05 m/s², r = 48.2 m

Substitute into equation 1

v = √(8.05×48.2)

v = √(388.01)

v = 19.698 m/s.

Hence the tangential speed of the car = 19.698 m/s

User Dinistro
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5.7k points