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Chapter 3: Problem 19

Challenge An object has an average acceleration of \(+6.24 \mathrm{~m} / \mathrm{s}^{2}\) for \(0.300 \mathrm{~s}\). At the end of this time the object's velocity is \(+9.31 \mathrm{~m} / \mathrm{s}\). What was the object's initial velocity?

Short Answer

Expert verified
The object's initial velocity was \(7.44 \, \mathrm{m/s}\).

Step by step solution

01

Understand the Given Information

We have an average acceleration \(a = +6.24 \ \mathrm{m/s^2}\), a time period \(t = 0.300 \ \mathrm{s}\), and a final velocity \(v_f = +9.31 \ \mathrm{m/s}\). We need to find the initial velocity \(v_i\).
02

Identify the Relevant Equation

The relationship between initial velocity \(v_i\), final velocity \(v_f\), acceleration \(a\), and time \(t\) is given by the equation: \[ v_f = v_i + a \cdot t \] This equation relates all the known variables and the unknown initial velocity.
03

Rearrange the Equation to Solve for Initial Velocity

We need to isolate \(v_i\) in the equation. We can do this by rearranging the equation: \[ v_i = v_f - a \cdot t \]
04

Substitute Known Values into the Equation

Substitute the given values into the rearranged equation: \[ v_i = 9.31 \, \mathrm{m/s} - (6.24 \, \mathrm{m/s^2} \times 0.300 \, \mathrm{s}) \] Calculating the product: \[ 6.24 \, \mathrm{m/s^2} \times 0.300 \, \mathrm{s} = 1.872 \, \mathrm{m/s} \]
05

Perform the Subtraction

Subtract the product from the final velocity to find the initial velocity: \[ v_i = 9.31 \, \mathrm{m/s} - 1.872 \, \mathrm{m/s} = 7.438 \, \mathrm{m/s} \]
06

Verify the Calculation

Verify the math and ensure all units are correctly canceled out: \(v_i = 7.438 \, \mathrm{m/s}\) is the correct initial velocity.

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Key Concepts

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

Initial Velocity
Initial velocity (\( v_i \)) is the velocity of an object at the start of a time interval. It's essential to determine changes in motion and calculate an object's trajectory. In this exercise, the task is to determine this initial velocity using known values of final velocity, acceleration, and time period.

When you know the final velocity (\( v_f \)), acceleration (\( a \)), and time interval (\( t \)), you can calculate the initial velocity by rearranging the kinematic equation:
  • \[ v_f = v_i + a imes t \]
By solving for \( v_i \), you get:
  • \[ v_i = v_f - a imes t \]
Understanding these concepts and knowing how to manipulate equations will help you solve similar problems effectively.
Final Velocity
Final velocity (\( v_f \)) represents the velocity of an object at the end of the observed time period. It is crucial in determining how much an object has accelerated or decelerated. In our exercise, we are given the final velocity, allowing us to work backwards to find the initial velocity.

The formula, \( v_f = v_i + a imes t \), shows that final velocity is a sum of initial velocity and the change in velocity caused by acceleration over time. With known values for final velocity, acceleration, and time, one can solve for initial velocity by rearranging this equation. Understanding final velocity helps predict the object's subsequent motion after a specific period.
Kinematic Equations
Kinematic equations are critical for solving motion problems, especially when dealing with constant acceleration. These equations relate displacement, velocity, acceleration, and time. They are valuable tools for predicting future motion or deducing past motion scenarios.

The key equations include:
  • \[ v_f = v_i + a imes t \]
  • \[ s = v_i imes t + \frac{1}{2}a imes t^2 \]
  • \[ v_f^2 = v_i^2 + 2a imes s \]
In our exercise, the first equation was used to find the initial velocity. Familiarity with each equation allows for flexible problem-solving strategies and helps tackle different types of questions in physics.
Time Period
The time period (\( t \)) is the duration over which motion is observed or measured. It's an integral part of motion equations, helping to quantify changes in an object's velocity due to acceleration. Knowing the time period is crucial in predicting how an object鈥檚 speed will change.

In the given exercise:
  • The time period is 0.300 seconds, allowing calculation of how much the object sped up from its initial velocity.
Using the equation \( v_i = v_f - a imes t \) showed the time period's role in determining initial velocity from a known final velocity. Understanding the time period鈥檚 effect on motion is key to successfully using kinematic equations.

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Most popular questions from this chapter

Surviving a Large Deceleration On July 13, 1977, while on a test drive at Britain's Silverstone racetrack, the throttle on David Purley's car stuck wide open. The resulting crash subjected Purley to the greatest \(g\)-force ever survived by a human-he decelerated from \(173 \mathrm{~km} / \mathrm{h}\) to zero in a distance of only about \(0.66 \mathrm{~m}\). Calculate the magnitude of the acceleration experienced by Purley (assuming it to be constant), and express your answer as a multiple of the acceleration due to gravity, \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

An object moves with constant acceleration. How is the average velocity of this object related to its initial and final velocities?

Think \& Calculate Assume that the brakes in your car create a constant deceleration of \(3.7 \mathrm{~m} / \mathrm{s}^{2}\) regardless of how fast you are driving. (a) If you double your driving speed from \(11 \mathrm{~m} / \mathrm{s}\) to \(22 \mathrm{~m} / \mathrm{s}\), does the distance required to come to a stop increase by a factor of 2 or a factor of 4? Explain. Verify your answer to part (a) by calculating the stopping distances for initial speeds of (b) \(11 \mathrm{~m} / \mathrm{s}\) and (c) \(22 \mathrm{~m} / \mathrm{s}\).

Triple Choice Truck 1 accelerates from \(5 \mathrm{~m} / \mathrm{s}\) to \(10 \mathrm{~m} / \mathrm{s}\) in \(10 \mathrm{~m}\). Truck 2 accelerates from \(15 \mathrm{~m} / \mathrm{s}\) to \(20 \mathrm{~m} / \mathrm{s}\) in \(10 \mathrm{~m}\). Is the acceleration of truck 1 greater than, less than, or equal to the acceleration of truck 2? Explain.

A car with an initial velocity of \(12 \mathrm{~m} / \mathrm{s}\) comes to rest in \(3.5 \mathrm{~s}\). What is the car's average acceleration during braking? Give both magnitude and sign.
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