A new car is tested on a 200-m-diameter track. If the car speeds up at a steady 1.5 m/s2, how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?

Ch 08: Dynamics II: Motion in a Plane

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 08: Dynamics II: Motion in a Plane

Problem 27

Chapter 8, Problem 27

A new car is tested on a 200-m-diameter track. If the car speeds up at a steady 1.5 m/s², how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?

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Step 1: Understand the relationship between centripetal acceleration and tangential acceleration. Centripetal acceleration is given by the formula \( a_c = \frac{v^2}{r} \), where \( v \) is the velocity and \( r \) is the radius of the circular path. Tangential acceleration is \( a_t = 1.5 \, \text{m/s}^2 \) (given in the problem). The condition for equality is \( a_c = a_t \).

Step 2: Substitute \( a_t \) and \( a_c \) into the equality condition. Using \( a_c = \frac{v^2}{r} \), set \( \frac{v^2}{r} = a_t \). The radius \( r \) of the track is half the diameter, so \( r = \frac{200}{2} = 100 \, \text{m} \). The equation becomes \( \frac{v^2}{100} = 1.5 \).

Step 3: Solve for \( v \) (velocity) in terms of the given tangential acceleration. Rearrange the equation \( \frac{v^2}{100} = 1.5 \) to \( v^2 = 1.5 \times 100 \). Then take the square root to find \( v \).

Step 4: Relate the velocity \( v \) to the time \( t \) using the formula for tangential acceleration. Since \( a_t = \frac{\Delta v}{\Delta t} \), and the car starts from rest, \( v = a_t \cdot t \). Substitute \( v \) from Step 3 into this equation to solve for \( t \).

Step 5: Perform the substitution and simplify. Replace \( v \) with \( \sqrt{1.5 \times 100} \) and \( a_t \) with \( 1.5 \) in the equation \( t = \frac{v}{a_t} \). This will give the time \( t \) when the centripetal acceleration equals the tangential acceleration.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Acceleration

Centripetal acceleration is the acceleration directed towards the center of a circular path that keeps an object moving in that path. It is calculated using the formula a_c = v^2 / r, where v is the tangential speed and r is the radius of the circular path. In this scenario, as the car speeds up, its tangential speed increases, affecting the centripetal acceleration.

Tangential Acceleration

Tangential acceleration refers to the rate of change of the speed of an object moving along a circular path. It is constant in this case, given as 1.5 m/s², and is responsible for increasing the car's speed as it moves around the track. This acceleration acts along the direction of the car's velocity, contributing to its overall motion.

Equating Accelerations

To find the time when centripetal acceleration equals tangential acceleration, we set the two expressions equal to each other. This involves using the formulas for both types of acceleration and solving for the time variable. The relationship between the car's speed, radius, and the given tangential acceleration is crucial for determining when these two accelerations are equal.

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