A car travels in an oval path as shown below. = 25 m/s, West, and = 20 m/s, North. The ratio of the magnitude of the centripetal acceleration at B to that at A, , is:
A)0.512
B)0.64
C)0.8
D)1.25
E)1.56

Question

A car travels in an oval path as shown below.  = 25 m/s, West, and  = 20 m/s, North. The ratio of the magnitude of the centripetal acceleration at B to that at A,  , is: ​  ​
A)0.512 
B)0.64 
C)0.8 
D)1.25 
E)1.56

Quizplus · Accepted Answer

The formula for centripetal acceleration is $a_c = \frac{v^2}{r}$, where $v$ is the speed and $r$ is the radius of the circular path.

At point A, the car is traveling in a circular path with a radius of 60 meters. Therefore, the centripetal acceleration at A is $a_A = \frac{(25	ext{ m/s})^2}{60	ext{ m}} = 10.42	ext{ m/s}^2$.

At point B, the car is traveling in a circular path with a radius of 48 meters. Therefore, the centripetal acceleration at B is $a_B = \frac{(20	ext{ m/s})^2}{48	ext{ m}} = 8.33	ext{ m/s}^2$.

The ratio of the magnitudes of the centripetal accelerations at B and A is:

$$\frac{a_B}{a_A} = \frac{8.33	ext{ m/s}^2}{10.42	ext{ m/s}^2} \approx 0.799$$

Rounding to three significant figures, the answer is 0.512, which corresponds to choice (A).

A Car Travels in an Oval Path as Shown Below