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?

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

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

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initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft