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Chapter 2: Problem 15

While driving on the freeway at \(110 \mathrm{km} / \mathrm{h},\) you pass a truck whose total length you estimate at 25 \(\mathrm{m}\) . (a) If it takes you, in the driver's seat, 5.5 \(\mathrm{s}\) to pass from the rear of the truck to its front, what is the truck's speed relative to the road? (b) How far does the truck travel while you're passing it?

Short Answer

Expert verified
(a) The truck's speed is approximately 93.64 km/h. (b) The truck travels approximately 143.06 m while you pass it.

Step by step solution

01

Identify the known quantities

We are given that your speed is \(110 \ \text{km/h}\), which we can convert to meters per second as \(110 \times \frac{1000}{3600} \ m/s\). The estimated length of the truck is \(25 \ \text{m}\), and the time it takes to pass the truck is \(5.5 \ \text{s}\).
02

Convert the speed of your car

Convert the speed from kilometers per hour to meters per second: \[110 \ \text{km/h} \times \frac{1000}{3600} \approx 30.56 \ \text{m/s}\]
03

Calculate relative speed of your car to the truck

The distance you cover while passing the truck, considering the truck's length, is the truck's length plus the distance the truck travels: \(25 \ \text{m}\). The time taken is \(5.5 \ \text{s}\). Use the formula: \[v_r = \frac{\text{distance}}{\text{time}} = \frac{25}{5.5} \]\[v_r \approx 4.55 \ \text{m/s}\]So, your speed relative to the truck is the car speed relative to the road minus the truck speed relative to the road. We have:\[30.56 \ \text{m/s} - v_t = 4.55 \ \text{m/s} \]
04

Solve for the truck's speed relative to the road

Rearrange the equation from Step 3 to solve for the truck's speed \(v_t\):\[v_t = 30.56 \ \text{m/s} - 4.55 \ \text{m/s} = 26.01 \ \text{m/s} \]Convert this back to km/h:\[v_t = 26.01 \times \frac{3600}{1000} \approx 93.64 \ \text{km/h}\]
05

Calculate the distance truck travels while you pass it

Since you pass the truck in \(5.5\) seconds, and the truck is traveling at \(26.01 \ \text{m/s}\), we use the formula for distance:\[\text{Distance} = v_t \times \text{time} = 26.01 \ \text{m/s} \times 5.5 \ \text{s} \approx 143.06 \ \text{m}\]

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Key Concepts

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

Kinematics
Kinematics is the study of motion without considering the forces causing the motion. In the exercise given, we explore concepts of speed and distance in a real-life scenario involving vehicles on a freeway. Understanding kinematics requires knowing how to describe the motion of objects through quantities such as velocity and acceleration.

When you are driving past a truck, your speed and the truck's speed pertain to kinematics. The problem here is to determine the relative speed between two objects moving in the same direction:
  • Your car's speed as observed from the ground.
  • The truck's speed as observed from the ground.
  • The truck's speed from your perspective inside the car.
Understanding these perspectives is key in kinematics and helps in quantifying motion in real-world scenarios.
Conversion of Units
Conversion of units is crucial in solving physics problems since different systems of units may be used for measuring similar quantities. In the problem, speed is initially given in kilometers per hour (km/h) and we need to convert it into meters per second (m/s) to perform calculations correctly.

To achieve this, use the conversion factor between these units:
  • 1 kilometer = 1000 meters,
  • 1 hour = 3600 seconds.

So, to convert the speed of 110 km/h to m/s, apply the formula: \[ 110 \ \text{km/h} \times \frac{1000}{3600} \approx 30.56 \ \text{m/s} \] This conversion allows us to consistently work in the metric system and easily apply kinematic equations to find the relative speeds and distances in the exercise.
Distance and Time Calculation
Calculating distance and time involves understanding and applying kinematic equations. To find out how far the truck travels while you pass it, you need its speed and the time it takes.

We already know from the exercise that the truck's speed relative to the road is 26.01 m/s and that you pass it in 5.5 seconds. Utilizing the formula for distance:
  • Distance = Speed 脳 Time,
we calculate: \[\text{Distance} = 26.01 \ \text{m/s} \times 5.5 \ \text{s} \approx 143.06 \ \text{m}\] Understanding this calculation helps us visualize real-life motion and relate it to measurable quantities of time and distance. It also demonstrates the practical use of the equation of motion, providing clarity on how these concepts interrelate to solve problems efficiently.

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Most popular questions from this chapter

An elite human sprinter reaches his top speed of 11.8 \(\mathrm{m} / \mathrm{s}\) at a time of 7.02 s after the starting gun. In the first \(1.40 \mathrm{s},\) however, he reaches a speed of \(8.00 \mathrm{m} / \mathrm{s},\) with a nearly constant acceleration. Calculate (a) his maximum acceleration during the starting phase and (b) his average acceleration to top speed. (c) Assuming constant acceleration for the first 1.40 s, how far does he travel during that time?

A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no appreciable air drag. (a) How tall, in meters, is the building? (b) How fast is the brick moving just before it reaches the ground? (c) Sketch graphs of this falling brick's acceleration, velocity, and vertical position as functions of time.

A mouse travels along a straight line; its distance \(x\) from the origin at any time \(t\) is given by the equation \(x=\left(8.5 \mathrm{cm} \cdot \mathrm{s}^{-1}\right) t-\left(2.5 \mathrm{cm} \cdot \mathrm{s}^{-2}\right) t^{2}\) . Find the average velocity of the mouse in the interval from \(t=0\) to \(t=1.0 \mathrm{s}\) and in the interval from \(t=0\) to \(t=4.0 \mathrm{s}\)

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of \(H,\) how high (in terms of \(H )\) will the faster stone go? Assume free fall.

(a) If a flea can jump straight up to a height of \(22.0 \mathrm{cm},\) what is its initial speed (in \(\mathrm{m} / \mathrm{s} )\) as it leaves the ground, neglecting air resistance? (b) How long is it in the air? (c) What are the magnitude and direction of its acceleration while it is (i) moving upward? (ii) moving downward? (iii) at the highest point?
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