/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 At a football tryout, a player r... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 21

At a football tryout, a player runs a 40 -yard dash in 4.25 seconds. If he reaches his maximum speed at the 16 -yard mark with a constant acceleration and then maintains that speed for the remainder of the run, determine his acceleration over the first 16 yards, his maximum speed, and the time duration of the acceleration.

Short Answer

Expert verified
Acceleration is approx 11.08 yards/sec虏, max speed approx 9.41 yards/sec, and acceleration lasts 1.7 sec.

Step by step solution

01

Analyzing the Problem

The player reaches his maximum speed after running 16 yards with constant acceleration. We need to determine three quantities: the acceleration during the first 16 yards, the maximum speed, and the time taken to reach that maximum speed.
02

Understanding Distance and Time Relationships

The total time for the 40-yard dash is 4.25 seconds. The player accelerates for the first 16 yards and then maintains a constant speed for the remaining 24 yards. Let \( t_1 \) be the time taken to reach the maximum speed, and \( t_2 \) be the time spent at maximum speed. Thus, \( t_1 + t_2 = 4.25 \) seconds.
03

Using Kinematics for the First 16 Yards

Since the player accelerates constantly over 16 yards, we use the equation \( d = \frac{1}{2}at_1^2 \), where \( d = 16 \) yards. Rearranging gives \( a = \frac{32}{t_1^2} \, \text{yards/sec}^2 \).
04

Calculating Maximum Speed

Maximum speed \( v \) can be computed as \( v = at_1 = 16/t_1 \, \text{yards/sec} \) using the acceleration formula from Step 3.
05

Total Time Calculation for 24 Yards

For the remaining 24 yards, the equation \( d = vt_2 \) gives \( 24 = v \times t_2 \). Substituting \( v = 16/t_1 \) into the equation gives \( t_2 = \frac{24t_1}{16} = \frac{3t_1}{2} \).
06

Solving the Total Time Equation

Using \( t_1 + t_2 = 4.25 \), we substitute \( t_2 = \frac{3t_1}{2} \) into the equation, leading to \( t_1 + \frac{3t_1}{2} = 4.25 \). Solving for \( t_1 \) gives \( \frac{5t_1}{2} = 4.25 \), hence \( t_1 = 1.7 \) seconds. Therefore, \( t_2 = 4.25 - 1.7 = 2.55 \) seconds.
07

Finding Acceleration

Using \( a = \frac{32}{t_1^2} \) and \( t_1 = 1.7 \), the acceleration is \( \approx 11.08 \, \text{yards/sec}^2 \).
08

Finding Maximum Speed

Plug \( t_1 = 1.7 \) into \( v = \frac{16}{t_1} \) to get the maximum speed, \( v \approx 9.41 \, \text{yards/sec} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration
When analyzing motion with constant acceleration, we understand that the object's velocity changes at a steady rate over time. In our exercise, the player experiences constant acceleration over the first 16 yards. To find the actual acceleration, we use the kinematic equation for distance:
  • \( d = \frac{1}{2} a t_1^2 \)
Inserting the known distance 16 yards, we adjust the formula to solve for acceleration \( a \):
  • \( a = \frac{32}{t_1^2} \)
Constant acceleration means the player steadily increases his speed until reaching a maximum speed at the end of this phase. Understanding this concept is key for solving problems related to velocity and time in kinematic exercises.
Maximum Speed
Maximum speed refers to the greatest velocity the player reaches during the run. This speed is first achieved at the 16-yard mark, where acceleration ceases and constant speed takes over.To calculate maximum speed using the kinematic relationships, we note:
  • Maximum speed \( v = a t_1 \)
  • Since \( a \) is determined by \( a = \frac{32}{t_1^2} \), then \( v = \frac{16}{t_1} \)
By substituting the values obtained from solving the time, the player's maximum speed at 16 yards is found. This is an essential trait as it influences the remaining distance and impacts the total time for such kinematic problems.
Time Duration
Determining the time duration is crucial since it tells us how long each part of the player's run lasts. Here, there are two specific durations:
  • \( t_1 \): Time to reach maximum speed (via acceleration over 16 yards)
  • \( t_2 \): Time spent maintaining maximum speed until the finish
From the total time equation:
  • \( t_1 + t_2 = 4.25 \, \text{seconds} \)
Using the relations between velocity and the distances, we solve for these individual times. Here:
  • \( t_1 = 1.7 \,\text{seconds} \)
  • \( t_2 = 2.55 \,\text{seconds} \)
This separation of time helps consider both acceleration and constant velocity phases, providing a full picture of the athlete's performance.
Distance-Time Relationship
The distance-time relationship is a cornerstone of kinematics, portraying how distance covered relates to time taken and speed. In the problem, the relationship is split into two segments:
  • 16 yards with constant acceleration
  • 24 yards at constant speed
First, the formula \( d = \frac{1}{2} a t_1^2 \) describes the path with acceleration, allowing us to solve for time and velocity. For the second phase with constant speed, the equation \( d = v t_2 \) applies. By combining these, we find out how each part contributes to the full journey's total time and verify the motion characteristics.Understanding how these segments work together ensures clarity in analyzing journeys with varying motion stages.

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Most popular questions from this chapter

Consider the polar axis of the earth to be fixed in space and compute the magnitudes of the velocity and acceleration of a point \(P\) on the earth's surface at latitude \(40^{\circ}\) north. The mean diameter of the earth is \(12742 \mathrm{km}\) and its angular velocity is \(0.7292\left(10^{-4}\right) \mathrm{rad} / \mathrm{s}\)

A toy helicopter is flying in a straight line at a constant speed of \(4.5 \mathrm{m} / \mathrm{s}\). If a projectile is launched vertically with an initial speed of \(v_{0}=28 \mathrm{m} / \mathrm{s}\), what horizontal distance \(d\) should the helicopter be from the launch site \(S\) if the projectile is to be traveling downward when it strikes the helicopter? Assume that the projectile travels only in the vertical direction.

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The rectangular coordinates of a particle are given in millimeters as functions of time \(t\) in seconds by \(x=30 \cos 2 t, y=40 \sin 2 t,\) and \(z=20 t+3 t^{2}\) Determine the angle \(\theta_{1}\) between the position vector \(\mathbf{r}\) and the velocity \(\mathbf{v}\) and the angle \(\theta_{2}\) between the position vector \(\mathbf{r}\) and the acceleration \(\mathbf{a},\) both at time \(t=2 \mathrm{s}\)
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