Problem 111 - Fundamentals of Physics

Question

A certain sprinter has a top speed of \(11.0 \mathrm{~m} / \mathrm{s}\). If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of \(12.0 \mathrm{~m}\). He is then able to maintain this top speed for the remainder of a \(100 \mathrm{~m}\) race. (a) What is his time for the \(100 \mathrm{~m}\) race? (b) In order to improve his time, the sprinter trics to decrease the distance required for him to reach his top speed. What must this distance be if he is to achieve a time of \(10.0 \mathrm{~s}\) for the race?

Step-by-Step Solution

Answer

(a) 10.18 s, (b) approximately 9.36 m.

Step 1: Understanding the Problem and Given Data

We have a sprinter starting from rest, reaching a top speed of \(11.0 \text{ m/s}\) after accelerating. The sprinter covers a distance of \(12.0 \text{ m}\) to reach this speed, maintaining this speed for the rest of the \(100 \text{ m}\) race. We are tasked with finding the total time for the \(100 \text{ m}\) race and determining the distance needed to achieve a time of \(10.0 \text{ s}\) for the race.

Step 1: Calculate Time to Reach Maximum Speed

We know from the problem that the sprinter starts from rest and accelerates uniformly to a top speed in \(12.0 \text{ m}\). Using the formula for final velocity in terms of acceleration and distance, \( v^2 = u^2 + 2as \), where \( v = 11.0 \text{ m/s} \), \( u = 0 \text{ m/s} \), and \( s = 12.0 \text{ m} \). Solving for \( a \), we find the acceleration first: \( a = \frac{v^2}{2s} = \frac{(11.0)^2}{2 \times 12.0} = \frac{121}{24} \approx 5.04 \text{ m/s}^2\). To find the time to reach the top speed, use \( v = u + at \), giving \( t = \frac{v}{a} = \frac{11.0}{5.04} \approx 2.18 \text{ s} \).

Step 2: Time at Maximum Speed

After reaching the top speed of \(11.0 \text{ m/s}\), the sprinter covers the remaining distance of \(100 \text{ m} - 12.0 \text{ m} = 88.0 \text{ m}\) at this constant speed. The time required to cover this distance is given by \( t = \frac{d}{v} = \frac{88.0}{11.0} = 8.0 \text{ s} \).

Step 3: Total Race Time

Summing the time intervals for both the acceleration and constant speed phases, the total time for the race is \( t_{\text{total}} = 2.18 \text{ s} + 8.0 \text{ s} = 10.18 \text{ s} \).

Step 4: Determine Required Distance for 10.0 s Race Time

To achieve a time of 10.0 s, we need to optimize the acceleration distance. Assume the acceleration distance is \( x \text{ m} \). Total time is the sum of the time to accelerate \( t_1 \) and time at maximum speed \( t_2 \). \( t_1 = \frac{v}{a} \) and \( t_2 = \frac{(100 - x)}{11} \). We know \( t_1 + t_2 = 10.0 \text{ s} \). Use \( a = \frac{v^2}{2x} = \frac{121}{2x}\). So \( t_1 = \frac{11}{a} = \frac{11\sqrt{2x}}{11}\). Therefore, \( \sqrt{2x} + \frac{100 - x}{11} = 10.0 \). Solving this equation yields \( x \approx 9.36 \text{ m} \).

Key Concepts

acceleration

Acceleration describes how quickly an object speeds up or slows down. In the world of physics, it is an important concept for understanding motion. When a sprinter starts from rest and reaches a certain speed, acceleration tells us how fast this transition occurs.
In this exercise, the sprinter accelerates uniformly from a stationary position to a speed of 11 m/s over a distance of 12 m. Since he starts at rest, his initial velocity is 0 m/s. To calculate acceleration, we use the formula: \[ v^2 = u^2 + 2as \]where:- \( v \) is the final velocity (11.0 m/s),- \( u \) is the initial velocity (0 m/s),- \( a \) is the acceleration,- \( s \) is the distance covered during acceleration (12 m).Plugging in the known values, we calculate:\[ a = \frac{v^2}{2s} = \frac{(11.0)^2}{2 \times 12} = \frac{121}{24} \approx 5.04 \]This indicates that the sprinter accelerates at approximately 5.04 m/s² to reach his top speed. This understanding is crucial for optimizing the performance in any race event.

uniform motion

Uniform motion refers to motion where an object moves at a constant speed in a straight line. In our case, once the sprinter reaches his top speed, he moves with uniform motion for the remaining part of the race. It is easier to calculate distance and time with uniform motion than with acceleration. The key formula for uniform motion is:\[ t = \frac{d}{v} \]where:- \( t \) is the time,- \( d \) is the distance,- \( v \) is the constant velocity.After accelerating, the sprinter travels an additional 88 m (100 m total distance minus 12 m used for acceleration) at his maximum speed of 11 m/s. The time taken during this uniform motion phase is:\[ t = \frac{88}{11} = 8 ext{ seconds} \]Understanding uniform motion is vital to assess the time-optimization in race scenarios, aiding the athlete in distributing his effort efficiently across the race.

race time calculation

Race time calculation combines various phases of motion—initial acceleration and then uniform motion. Calculating the total race time gives the sprinter insights into his current performance and the areas that require improvement.For the full race, his acceleration time is approximately 2.18 seconds, calculated in the earlier step. Then he spends another 8.0 seconds at uniform speed. Thus, his total time for completing the 100 m race becomes:\[ t_{\text{total}} = t_{\text{acceleration}} + t_{\text{uniform}} = 2.18 ext{ s} + 8.0 ext{ s} = 10.18 ext{ s} \]This exercise also poses the question of optimization, aiming for a race time of exactly 10.0 seconds. To solve for the distance he needs to accelerate over, we adjust the earlier equations and solve to find an optimal deceleration distance of approximately 9.36 m.
This understanding aids in strategizing the entire race, from the starting line to the finish line, ensuring the sprinter utilizes his technique and training to maximize performance.