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In getting ready to slam-dunk the ball, a basketball player starts from rest and sprints to a speed of \(6.0 \mathrm{m} / \mathrm{s}\) in \(1.5 \mathrm{s}\). Assuming that the player accelerates uniformly, determine the distance he runs.

Short Answer

Expert verified
The player runs 4.5 meters.

Step by step solution

01

Identify the Known Variables

We need to list the known quantities: the initial speed \( v_i = 0 \) (starts from rest), the final speed \( v_f = 6.0 \, \text{m/s} \), and the time taken \( t = 1.5 \, \text{s} \).
02

Write Down the Equation for Uniform Acceleration

The formula for distance \( d \) covered under uniform acceleration is given by:\[ d = v_i \cdot t + \frac{1}{2} a \cdot t^2 \]We already know \( v_i \) and \( t \), but we need to find acceleration \( a \).
03

Find the Acceleration

Using the formula for acceleration under uniform velocity change:\[ a = \frac{v_f - v_i}{t} \]Substitute the known values:\[ a = \frac{6.0 \, \text{m/s} - 0}{1.5 \, \text{s}} = 4.0 \, \text{m/s}^2 \]
04

Substitute Values into the Distance Formula

Now use the calculated acceleration value in the distance formula:\[ d = 0 \cdot 1.5 + \frac{1}{2} \cdot 4.0 \, \text{m/s}^2 \cdot (1.5 \, \text{s})^2 \]This simplifies to:\[ d = 0 + 0.5 \cdot 4.0 \cdot 2.25 = 4.5 \, \text{m} \]
05

Conclusion

The basketball player runs a total distance of \( 4.5 \, \text{m} \) while accelerating to the slam dunk.

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Key Concepts

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

Kinematics
Kinematics is a branch of mechanics that focuses on the motion of objects without considering the forces that cause this motion. In simpler terms, it's all about understanding how something moves from one point to another. Various variables like speed, distance, time, and acceleration are used in kinematics to describe motion.
For instance, when solving motion problems, it's important to determine what is given (like an object's speed or time) and what needs to be found (such as distance traveled).
This helps in using the right kinematic equations, which are mathematical expressions that describe motion based on these variables.
  • The three fundamental kinematic equations relate speed, distance, acceleration, and time.
  • Kinematics helps in predicting future positions of objects and in understanding their past motions.
Understanding kinematics provides a base for tackling various physical scenarios efficiently.
Initial Speed
Initial speed, often denoted as \( v_i \), is the speed at which an object starts its motion. It is a crucial variable in kinematic calculations as it sets the starting condition for any motion analysis.
The initial speed can be anything, from zero (an object at rest) to several meters per second, depending on the problem context.
In the exercise, the basketball player starts from rest, which means the initial speed is 0 \( ext{m/s} \).
  • Initial speed is important for calculating distance and determining changes in motion.
  • When an object starts from rest, calculations often become simpler because \( v_i = 0 \).
Whether an object is accelerating or decelerating, the initial speed allows one to understand the starting point of its movement.
Final Speed
Final speed, symbolized as \( v_f \), represents the speed of an object at the end of a given period. It's the speed after the object has experienced acceleration or deceleration.
Determining the final speed helps in understanding how quickly an object is moving after a set time or event has occurred.
In the case of the basketball player, his final speed is 6.0 \( ext{m/s} \).
  • Final speed can be found using various kinematic equations.
  • It's a key determinant in problems involving acceleration, as it reflects the change from initial speed.
Knowing the final speed is essential to assess performance, such as how fast an athlete runs or a vehicle drives after starting off.
Acceleration Formula
Acceleration is the rate at which an object speeds up or slows down, and it is calculated using a straightforward formula. It is denoted as \( a \), and the formula in uniform acceleration cases is: \[ a = \frac{v_f - v_i}{t} \]Where \( v_f \) is the final speed, \( v_i \) is the initial speed, and \( t \) is the time over which the acceleration occurs.
The exercise demonstrates the calculation needed to find acceleration.
  • In the example, \( a = \frac{6.0 \, \text{m/s} - 0}{1.5 \, \text{s}} = 4.0 \, \text{m/s}^2 \).
  • Acceleration tells us how quickly the player increased his speed to prepare for the slam dunk.
Uniform acceleration assumes no changes in force or direction, simplifying calculations and predictions.

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Most popular questions from this chapter

A woman and her dog are out for a morning run to the river, which is located \(4.0 \mathrm{km}\) away. The woman runs at \(2.5 \mathrm{m} / \mathrm{s}\) in a straight line. The dog is unleashed and runs back and forth at \(4.5 \mathrm{m} / \mathrm{s}\) between his owner and the river, until the woman reaches the river. What is the total distance run by the dog?

A jet is taking off from the deck of an aircraft carrier, as shown in the image. Starting from rest, the jet is catapulted with a constant acceleration of \(+31 \mathrm{m} / \mathrm{s}^{2}\) along a straight line and reaches a velocity of \(+62 \mathrm{m} / \mathrm{s} .\) Find the displacement of the jet.

A VW Beetle goes from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) with an acceleration of \(+2.35 \mathrm{m} / \mathrm{s}^{2}\) (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to \(60.0 \mathrm{mi} / \mathrm{h}\) in \(0.600 \mathrm{s}\). Find the acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) of the dragster.

A diver springs upward with an initial speed of \(1.8 \mathrm{m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. / Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{m}\) (measured from the board), assuming that the dowmward direction is chosen as the negative direction./ (b) What is the highest point he reaches above the water?

A police car is traveling at a velocity of \(18.0 \mathrm{m} / \mathrm{s}\) due north, when a car zooms by at a constant velocity of \(42.0 \mathrm{m} / \mathrm{s}\) due north. After a reaction time of \(0.800 \mathrm{s}\) the policeman begins to pursue the speeder with an acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2} .\) Including the reaction time, how long does it take for the police car to catch up with the speeder?

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