Problem 74 - College Physics

Question

A car accelerates uniformly from rest, and is initially pointed north. It then travels in a quarter circle taking \(10.0 \mathrm{~s}\) and reaching a final speed of \(30.0 \mathrm{~m} / \mathrm{s}\) traveling due east. (a) What is the radius of its path? At 5.0 s from the start, determine (b) the car's tangential acceleration (including direction), \((\) c) the car's centripetal acceleration (including direction), and (d) the car's velocity.

Step-by-Step Solution

Answer

Path radius is 29.8 m. Tangential acceleration at 5 s is 3 m/s² east. Centripetal acceleration at 5 s is 7.54 m/s². Velocity at 5 s is 15 m/s at 45°.

Step 1: Understand the Problem

A car accelerates uniformly from rest along a quarter circle path, taking 10.0 seconds to reach a velocity of 30.0 m/s. We need to find the radius of the path, and at 5.0 seconds, determine the car's tangential acceleration, centripetal acceleration, and velocity.

Step 2: Calculate the Radius of the Path

Using the concept of uniform circular motion and acceleration, we start by using the equation for velocity: \( v = r \omega \). Since it changes direction into a quarter circle path, it implies \( v_{f}^2 = v_{i}^2 + 2ar\). Given \( v_i = 0 \), solve for \( r \): \[ 30.0 = \sqrt{0 + 2 \times a \cdot r} \] Rearrange and plug in values: \( r = \frac{v^2}{2a}\), \( 30^2 = 2ar \rightarrow a = \frac{900}{2r} \), applying \( s = r\theta \). But knowing \( s = r \cdot \frac{\pi}{2} = vt \rightarrow 10r = \frac{30 \times \pi}{10} \rightarrow r = 29.8 \text{ m} .\)

Step 3: Calculate the Tangential Acceleration at 5 s

Using the relationship between velocity and acceleration, \( a_t = \frac{v_f - v_i}{t} = \frac{30.0 \text{ m/s} - 0 \text{ m/s}}{10.0 \text{ s}} = 3.0 \text{ m/s}^2 \). This is directed along the tangent to the path. At \( 5 \text{ s} \), the car's tangential acceleration is still \( 3.0 \text{ m/s}^2 \) toward east.

Step 4: Calculate the Centripetal Acceleration at 5 s

Centripetal acceleration \( a_c = \frac{v^2}{r} \). At \( 5 \text{ s}\), it's essential to find the velocity: \( v = 3.0 \times 5 = 15 \, \text{m/s} \). Substitute in \( a_c = \frac{15^2}{29.8} \approx 7.54 \text{ m/s}^2\). This acts perpendicular to the velocity, toward the center of the circular path.

Step 5: Calculate Velocity at 5 s

The velocity at 5 seconds, using \( v = v_i + a_t \times t \), initially \( v_i = 0\). \( v = 0 + 3.0 \time 5 = 15 \text{ m/s} \). Since the car is at 45 degrees into its turn, the velocity has components: \( v_x = v_y = \frac{15}{\sqrt{2}} \).

Step 6: Result Verification and Conclusion

Summarizing, we computed: \(\text{(a)}\) Path radius \(\approx 29.8\) meters, \(\text{(b)}\) Tangential acceleration \(3.0\text{ m/s}^2 \) east, \(\text{(c)}\) Centripetal acceleration \(\approx 7.54 \text{ m/s}^2\) towards center, \(\text{(d)}\) Velocity \(15 \text{ m/s}\) at 45 degrees.

Key Concepts

Tangential Acceleration

When a car moves along a curved path like a quarter-circle, it experiences tangential acceleration which affects its speed along the path. This kind of acceleration is responsible for speeding up or slowing down the car as it travels. In our exercise, the car starts from rest, meaning its initial speed is 0 m/s and reaches a speed of 30 m/s over a period of 10 seconds.

• Formula: The tangential acceleration can be calculated using the formula: \( a_t = \frac{v_f - v_i}{t} \), where \( v_f \) is the final speed, \( v_i \) is the initial speed, and \( t \) is the time taken.
• Example Calculation: Plugging in our values, we find: \( a_t = \frac{30.0 - 0}{10.0} = 3.0 \, \text{m/s}^2 \). This means that the car's speed changes by 3.0 m/s every second along the path.
• Direction: Since the car travels east at 5 seconds, this tangential acceleration is directed along the tangent of the curved path in the eastward direction.

Understanding tangential acceleration helps to understand why the car speeds up smoothly as it moves along a curved path.

Centripetal Acceleration

Centripetal acceleration is crucial in circular motion as it keeps an object on its curved path. It acts perpendicular to the object's instantaneous velocity and points toward the center of the circle. This is what prevents the car from moving in a straight line off the path.

• Formula: The centripetal acceleration is given by the formula: \( a_c = \frac{v^2}{r} \), where \( v \) is the velocity of the object, and \( r \) is the radius of the circle.
• Example Calculation: At 5 seconds into the journey, the car's velocity is 15 m/s. The radius of the path was calculated earlier as 29.8 m. Using the formula, \( a_c = \frac{15^2}{29.8} \approx 7.54 \, \text{m/s}^2 \). This indicates the force pulling the car inward to maintain its curved path.
• Direction: This acceleration acts perpendicular to the car's path towards the center of the circle.

Centripetal acceleration is what ensures the car stays on track instead of veering off due to inertia.

Velocity Calculation

Velocity in circular motion combines speed and direction. In this context, as the car moves through a quarter-circle, its direction changes constantly. Calculating velocity at a specific point in the path provides insight into this changing direction.

• Formula: The relationship is expressed as: \( v = v_i + a_t \times t \).
• At 5 Seconds: Initially, the car's velocity was 0 m/s since it started from rest. At 5 seconds \( v = 0 + 3.0 \times 5 = 15 \text{ m/s} \). This only determines the speed, so we have to understand direction as well.
• Components: At 5 seconds into a quarter-circle turn, the car has traveled 45 degrees from north, meaning its velocity is equally distributed between east and north: \( v_x = v_y = \frac{15}{\sqrt{2}} \).

The calculation of velocity helps us understand not just how fast the car is moving, but also in which direction it is pointing at any point in its circular path.