Question

A car travels at a constant speed around a circular track whose radius is 3.90 km. The car goes once around the track in 292 s. What is the magnitude of the centripetal acceleration of the car?

Answer 1

Answer on Question #49036, Physics, Mechanics | Kinematics | Dynamics

A car travels at a constant speed around a circular track whose radius is $3.90 \, \text{km}$ . The car goes once around the track in 292 s. What is the magnitude of the centripetal acceleration of the car?

Solution.

By definition:

$$a _ {C} = \frac {v ^ {2}}{R}$$

Speed of the car is equal to length of the circle divided by time of one revolution:

$$v = \frac {2 \pi R}{\mathrm {T}}$$

So:

$$a _ {C} = \frac {v ^ {2}}{R} = \left(\frac {2 \pi R}{\mathrm {T}}\right) ^ {2} \frac {1}{R} = \frac {4 \pi^ {2} R}{\mathrm {T} ^ {2}}$$

Numerically:

$$a _ {C} = \frac {4 \cdot 3 . 1 4 ^ {2} \cdot 3 . 9 0 \cdot 1 0 ^ {3} m}{(2 9 2 \mathrm {s}) ^ {2}} \approx 1. 8 \frac {m}{s ^ {2}}$$

Answer: $a_{C} = 1.8\frac{m}{s^{2}}$

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