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A Fast Pitch. The fastest measured pitched baseball left the pitcher's hand at a speed of 45.0 \(\mathrm{m} / \mathrm{s}\) . If the pitcher was in contact with the ball over a distance of 1.50 \(\mathrm{m}\) and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?

Short Answer

Expert verified
(a) The acceleration is 675 m/s². (b) The time taken is 0.0667 s.

Step by step solution

01

Identify the Given Values

The problem provides the initial speed of the ball as 0 m/s (since it starts from rest) and the final speed as 45.0 m/s. The distance over which the ball was accelerated by the pitcher is given as 1.50 m. We need to find the acceleration and the time taken for the pitch.
02

Use the Kinematic Equation for Constant Acceleration

We will use the kinematic equation: \[ v^2 = u^2 + 2as \]where:- \( v = 45.0 \, \text{m/s} \) is the final velocity,- \( u = 0 \, \text{m/s} \) is the initial velocity,- \( a \) is the acceleration,- \( s = 1.50 \, \text{m} \) is the distance.Substituting the known values:\[ (45.0)^2 = 0 + 2a(1.50) \]\[ 2025 = 3a \]
03

Calculate the Acceleration

Solve for \( a \): \[ a = \frac{2025}{3} = 675 \, \text{m/s}^2 \]The acceleration given to the ball is \( 675 \, \text{m/s}^2 \).
04

Use the Kinematic Equation to Find Time

Now, we use the equation:\[ v = u + at \]where:- \( v = 45.0 \, \text{m/s} \)- \( u = 0 \, \text{m/s} \)- \( a = 675 \, \text{m/s}^2 \)Rearranging to solve for time \( t \):\[ t = \frac{v - u}{a} = \frac{45.0}{675} \]
05

Calculate the Time

Calculate:\[ t = \frac{45.0}{675} = 0.0667 \, \text{s} \]The time taken for the pitcher to accelerate the ball over the given distance is approximately \( 0.0667 \, \text{s} \).

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Key Concepts

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

Kinematic Equations
Kinematic equations are pivotal in physics for analyzing motion, particularly when acceleration is constant. These equations relate the five primary components of motion: initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), time (\(t\)), and displacement (\(s\)). Each kinematic equation connects certain combinations of these variables.

In the given exercise, the focus is on finding acceleration and time using specific kinematic equations. The equation \(v^2 = u^2 + 2as\) connects final velocity, initial velocity, acceleration, and displacement. This equation helped determine how the pitcher's acceleration affected the baseball.

The second equation, \(v = u + at\), is used to calculate the time it takes for the velocity to change given a constant acceleration. Leveraging these equations allows us to solve problems where direct measurements of acceleration and time might not be feasible.

Understanding when and how to apply these equations is an important problem-solving skill and the key to dissecting motion scenarios effectively.
Constant Acceleration
Constant acceleration implies that the velocity of an object increases by the same amount every second. In the context of the given baseball problem, the pitcher imparts a uniform increase in speed over the distance of 1.50 meters, ensuring that the ball undergoes constant acceleration.

This concept simplifies calculations significantly because the formulas for kinematic equations depend on acceleration being consistent. When dealing with constant acceleration, the average velocity during intervals remains straightforward to calculate since it forms a predictable path. In scenarios like the baseball pitch, knowing the constant acceleration allows us to directly compute the other motion factors, avoiding complications arising from varying acceleration.

It's important to note that constant acceleration may not always occur in real-life situations. However, for many simpler physics problems and preliminary calculations, this assumption helps streamline the analysis and understanding of motion.
Problem-solving Steps
Solving physics problems is often about following a structured approach to ensure all factors are considerately analyzed. The step-by-step method begins with identifying known variables and what needs to be determined. Recognizing these elements early on sets up the pathway to selecting appropriate equations.

Next, deploying relevant kinematic equations facilitates finding unknown variables, as demonstrated in the baseball pitch problem. Using the equation \(v^2 = u^2 + 2as\) allowed us to isolate and compute acceleration. Similarly, solving another kinematic equation yielded the time taken during the pitch.

This problem-solving framework involves calculating each phase thoughtfully, double-checking calculations to ensure logic flows and equations align correctly. By systematically breaking down the problem, complex physics becomes more approachable and solvable one step at a time.

Approach every physics problem as an opportunity to practice these steps, refine your technique, and gain deeper insights into motion and its governing laws.

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Most popular questions from this chapter

Trip Home. You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 \(\mathrm{km} / \mathrm{h}\) \((65 \mathrm{mi} / \mathrm{h}),\) and the trip takes 2 \(\mathrm{h}\) and 20 \(\mathrm{min}\) . On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 \(\mathrm{km} / \mathrm{h}\) (43 milh). How much longer does the trip take?

On a 20 mile bike ride, you ride the first 10 miles at an average speed of 8 \(\mathrm{mi} / \mathrm{h}\) . What must your average speed over the next 10 miles be to have your average speed for the total 20 miles be (a) 4 \(\mathrm{mi} / \mathrm{h} ?\) (b) 12 \(\mathrm{mi} / \mathrm{h} ?(\mathrm{c})\) Given this average speed for the first 10 miles, can you possibly attain an average speed of 16 \(\mathrm{mi} / \mathrm{h}\) for the total 20 -mile ride? Explain.

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2} .\) Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 \(\mathrm{s}\) , Powers shuts off the engine and steps out of the helicopter Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 \(\mathrm{m} / \mathrm{s}^{2} .\) How far is Powers above the ground when the helicopter crashes into the ground?

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.20 \(\mathrm{m} / \mathrm{s}^{2} .\) At the same instant a truck, traveling with a constant speed of 20.0 \(\mathrm{m} / \mathrm{s}\) , overtakes and passes the car. (a) How far beyond its starting point does the car overtake the truck? (b) How fast is the car traveling when it overtakes the truck? (c) Sketch an \(x\) -t graph of the motion of both vehicles. Take \(x=0\) at the inter-section. \(\left(\text { d) Sketch a } v_{x}-t \text { graph of the motion of both vehicles. }\right.\)

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 \(\mathrm{s}\) . It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

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