Chapter 2: Problem 87
In a quarter-mile drag race, two cars start simultaneously from rest, and each accelerates at a constant rate until it either reaches its maximum speed or crosses the finish line. Car A has an acceleration of 11.0 \(\mathrm{m} / \mathrm{s}^{2}\) and a maximum speed of 106 \(\mathrm{m} / \mathrm{s}\) . Car \(\mathrm{B}\) has an acceleration of 11.6 \(\mathrm{m} / \mathrm{s}^{2}\) and a maximum speed of 92.4 \(\mathrm{m} / \mathrm{s}\) . Which car wins the race, and by how many seconds?
Short Answer
Step by step solution
Convert Distance
Calculate Time for Car A to Reach Maximum Speed
Calculate Distance Covered by Car A to Reach Maximum Speed
Calculate the Time for Car A to Finish the Race
Calculate Time for Car B to Reach Maximum Speed
Calculate Distance Covered by Car B to Reach Maximum Speed
Calculate the Time for Car B to Finish the Race
Determine the Winner and Time Difference
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Constant Acceleration
- \( v = u + at \)
- \( v \) is the final velocity
- \( u \) is the initial velocity (usually 0 in these problems)
- \( a \) is the acceleration
- \( t \) is time
Quarter Mile to Meters
- \( ext{Quarter mile in meters} = \frac{1609}{4} \approx 402.25 \text{ meters} \)
Maximum Speed
- \( t = \frac{v}{a} \)
- \( v \) is the maximum speed
- \( a \) is the acceleration
Time Calculation
- Using the formula \( x = \frac{1}{2} a t^2 \)
- Solving for \( t \) gives \( t = \sqrt{\frac{2x}{a}} \)
- \( x \) is the distance (in meters)
- \( a \) is acceleration
- \( t \) is the time to be calculated