Question

The driver of a car moving at 90.0 km/h presses down on the brake as the car enters a circular curve of radius 195.0 m. If the speed of the car is decreasing at a rate of 4.0 km/h each second, what is the magnitude of the acceleration of the car (in m/s2) at the instant its speed is 58.0 km/h?

Answer 1

Answer:

Approximately $\rm 1.73 \; \rm m \cdot s^{-2}$.

Explanation:

The acceleration on the car comes in two parts:

• The acceleration is due to the brakes is tangential to the circular curve.
• The acceleration due to the circular motion of the car is centripetal. It points towards the center of the circle.

These two accelerations are perpendicular to each other.

Convert the current velocity to meters-per-second ($\rm m \cdot s^{-1}$.)

$\begin{aligned} v &= 58.0\; \rm km \cdot h^{-1} \\ &= \left(58.0\; \rm km \cdot h^{-1}\left/\frac{3600\; \rm s}{1\; \rm h}\right.\right) \times \frac{1000\; \rm m}{1 \; \rm km} \\ &= \frac{58.0}{3600} \times 1000 \; \rm m \cdot s^{-1} \\ & \approx 16.1\; \rm m \cdot s^{-1}\end{aligned}$.

Convert the acceleration due to braking to meters-per-second-squared ($\rm m \cdot s^{-2}$.)

$\begin{aligned}a_{\text{tangential}} &= 4.0\; \rm km \cdot h^{-1} \cdot s^{-1} \\ &= \left(4.0\; \rm km \cdot h^{-1} \cdot s^{-1}\left/\frac{3600\; \rm s}{1\; \rm h}\right.\right) \times \frac{1000\; \rm m}{1 \; \rm km} \\ &= \frac{4.0}{3600} \times 1000 \; \rm m \cdot s^{-1} \cdot s^{-1} \\ & \approx 1.111111\; \rm m \cdot s^{-2} \end{aligned}$.

Calculate the acceleration due to the circular motion:

$\begin{aligned} a_\text{centripetal} &= \frac{v^2}{r} \\ &= \frac{\left(16.1\; \rm m \cdot s^{-1}\right)^2}{195.0\; \rm m} \\ & \approx 1.33112\; \rm m \cdot s^{-2}\end{aligned}$.

Since the two accelerations are perpendicular to each other, the resultant acceleration can be found using the Pythagorean Theorem

$\begin{aligned}a &= \sqrt{\left(a_\text{tangential}\right)^2 + \left(a_\text{centripetal}\right)^2} \\ &\approx \sqrt{1.111111^2 + 1.33112^2} \\ &\approx 1.73\; \rm m \cdot s^{-2}\end{aligned}$.