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Chapter 4: Problem 81

Distance and speed. A motorcycle accelerates at a constant rate from 0 mph \((v(0)=0 \text { ) to } 60 \text { mph in } 15\) sec. a) How fast is it traveling after 15 sec? b) How far has it traveled after 15 sec? (Hint: Convert seconds to hours.)

Short Answer

Expert verified
a) 60 mph b) 0.125 miles

Step by step solution

01

Understand the given values and units

The motorcycle accelerates from 0 mph to 60 mph in 15 seconds. Note that speed is given in mph but time is in seconds.
02

Calculate acceleration

Acceleration is the change in velocity over time. Given the initial velocity is 0 mph, the final velocity is 60 mph, and the time is 15 seconds, the formula for constant acceleration is: \[ a = \frac{\text{final velocity} - \text{initial velocity}}{\text{time}} \]Convert 15 seconds to hours: \[ 15 \text{ sec} = \frac{15}{3600} \text{ hours} = \frac{1}{240} \text{ hours} \]Now compute the acceleration: \[ a = \frac{60 \text{ mph} - 0 \text{ mph}}{\frac{1}{240} \text{ hours}} = 60 \times 240 = 14400 \text{ mph}^2 \]
03

Calculate the speed after 15 seconds

Using the acceleration, the speed after 15 seconds (which is also the endpoint) is 60 mph, since the problem states the motorcycle reaches this speed.
04

Calculate distance traveled

To find the distance traveled, use the formula for distance traveled under constant acceleration from rest: \[ d = \frac{1}{2} a t^2 \]First convert time 15 seconds to hours again: \[ t = \frac{15}{3600} \text{ hours} = \frac{1}{240} \text{ hours} \]Now substitute the values into the distance formula: \[ d = \frac{1}{2} \times 14400 \text{ mph}^2 \times \left(\frac{1}{240} \text{ hours}\right)^2 \] Simplify this step-by-step: \[ d = 7200 \times \left(\frac{1}{240} \right)^2 = 7200 \times \frac{1}{57600} = \frac{7200}{57600} = \frac{1}{8} \text{ miles} = 0.125 \text{ miles} \]

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Key Concepts

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

Distance Calculation
To figure out how far a motorcycle has traveled, use the distance formula for an object with constant acceleration from rest: \(d = \frac{1}{2} a t^2 \). This formula helps determine the distance over time.
First, it's crucial to convert any time given in seconds to hours if the velocity is measured in miles per hour.
In our example, we converted 15 seconds to hours: \(\frac{15}{3600} = \frac{1}{240}\) hours. Next, we applied the acceleration value and time to calculate the distance as follows:
Given \(a = 14400 \text{ mph}^2\) and \(t = \frac{1}{240} \text{ hours}\), the distance \(d\) is:
\(d = \frac{1}{2} \times 14400 \times \left(\frac{1}{240}\right)^2 = 7200 \times \frac{1}{57600} = \frac{7200}{57600} = \frac{1}{8} \text{ miles} = 0.125 \text{ miles}\). This conversion and substitution step is vital to ensure values align correctly for accurate distance calculation.
Velocity
Velocity is a measure of the speed of an object in a particular direction. In this exercise, we determine how fast the motorcycle is traveling after 15 seconds. The motorcycle accelerates from 0 mph to 60 mph.
Using the formula for constant acceleration: \(a = \frac{\text{final velocity} - \text{initial velocity}}{\text{time}} \), we can calculate the acceleration. Here, the initial velocity \(v(0) = 0 \text{ mph}\) and final velocity \(v(15 \text{ sec}) = 60 \text{ mph}\), with time \(t = 15 \text{ sec}\) converted to hours \( \frac{1}{240} \text{ hours}\):
\(a = \frac{60 - 0}{\frac{1}{240}} = 60 \times 240 = 14400 \text{ mph}^2\).
With constant acceleration, the motorcycle reaches 60 mph after 15 seconds.
Time Conversion
Converting time units correctly is crucial when solving physics problems. In our scenario, speed is given in miles per hour (mph), while time is in seconds.
To ensure consistency, we need to convert seconds to hours: \(1 \text{ hour} = 3600 \text{ seconds}\). For 15 seconds: \( \frac{15}{3600} = \frac{1}{240} \text{ hours}\).
This conversion allows us to apply formulas using compatible units. Always check the units of measure when performing calculations, and convert where necessary to avoid errors and ensure accurate results.

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