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Chapter 2: Problem 27

A car traveling east at \(40.0 \mathrm{~m} / \mathrm{s}\) passes a trooper hiding at the roadside. The driver uniformly reduces his speed to \(25.0 \mathrm{~m} / \mathrm{s}\) in \(3.50 \mathrm{~s}\). (a) What is the magnitude and direction of the car's acceleration as it slows down? (b) How far does the car travel in the \(3.5-\mathrm{s}\) time period?

Short Answer

Expert verified
The magnitude of the car's acceleration as it slows down is \(4.29 \, m/s^2\) directed west, or in the opposite direction to the car鈥檚 motion, and the car travels \(98.5 \, m\) in the \(3.50 \, s.\)

Step by step solution

01

Identify Given Values

In the problem, the following values are given: initial speed (\(v_i = 40.0 \, m/s\)), final speed (\(v_f = 25.0 \, m/s\)), and time (\(t = 3.50 \, s\)).
02

Calculate Acceleration

We use the formula for acceleration, \(a = (v_f - v_i) / t\). Substituting the given values, we have \(a = (25.0 \, m/s - 40.0 \, m/s) / 3.50 \, s = -4.29 \, m/s^2\). The negative sign indicates that the acceleration is directed west, or in the opposite direction to the car鈥檚 motion, because the car is slowing down.
03

Calculate Distance

The distance covered by the car can be calculated using the formula \(d = v_i t + 0.5 a t^2\). Substituting in the coordinates we get: \(d = 40.0 \, m/s * 3.50 \, s + 0.5 * -4.29 \, m/s^2 * (3.50 \, s)^2 = 98.5 \, m\).

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Key Concepts

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

Understanding Acceleration
Acceleration tells us how quickly an object changes its velocity. It is a key concept in kinematics, helping us grasp how fast or slow something speeds up or slows down.
To calculate acceleration, you can use the formula: \(a = \frac{v_f - v_i}{t}\).
  • Here, \(v_i\) is the initial velocity.
  • \(v_f\) is the final velocity.
  • \(t\) is the time it takes for the change in velocity.
In our scenario, the car's initial speed was \(40 \ m/s\) east, and it slowed to \(25 \ m/s\) over \(3.5 \, s\). This yielded an acceleration of \(-4.29 \, m/s^2\).
The negative sign indicates the acceleration is in the opposite direction to the motion, meaning the car is decelerating. Understanding the direction is important in recognizing why the car slows down. This concept is crucial for analyzing real-world motions.
Calculating Distance Traveled
Distance traveled refers to the length of the path an object takes as it moves over a period of time. In kinematics, to find this for an object moving with uniform acceleration, we use the equation: \(d = v_i \times t + 0.5 \times a \times t^2\).
  • The term \(v_i \times t\) accounts for the distance covered if the velocity remained constant.
  • The term \(0.5 \times a \times t^2\) adjusts for the change due to acceleration.
For our car, traveling with an initial velocity of \(40 \, m/s\) and an acceleration of \(-4.29 \, m/s^2\) for \(3.5 \, s\), the formula shows it covers \(98.5 \, m\).
Recognizing the components of this formula helps us understand how acceleration influences different types of motion, whether speeding up or slowing down.
Exploring Velocity
Velocity is a measure of how fast an object moves and in which direction. Unlike speed, velocity considers direction, making it a vector quantity.
Initially, our car had a velocity of \(40.0 \ m/s\) towards the east. As it slowed down, the velocity decreased to \(25.0 \ m/s\) in the same direction.
  • The change in velocity here helps us calculate acceleration.
  • The direction of velocity is crucial, especially when calculating acceleration and distance.
Understanding the change from higher to lower velocity helps determine how fast an object slows down. In other contexts, velocity changes might involve different directions, requiring careful consideration of vector components.
Through this kind of analysis, one sees how velocity intertwines with other kinematic concepts to help us understand motion comprehensively.

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Most popular questions from this chapter

An attacker at the base of a castle wall \(3.65 \mathrm{~m}\) high throws a rock straight up with speed \(7.40 \mathrm{~m} / \mathrm{s}\) at a height of \(1.55 \mathrm{~m}\) above the ground. (a) Will the rock reach the top of the wall? (b) If so, what is the rock's speed at the top? If not, what initial speed must the rock have to reach the top? (c) Find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of \(7.40 \mathrm{~m} / \mathrm{s}\) and moving between the same two points. (d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Explain physically why or why not.

A parachutist with a camera descends in free fall at a speed of \(10 \mathrm{~m} / \mathrm{s}\). The parachutist releases the camera at an altitude of \(50 \mathrm{~m}\). (a) How long does it take the camera to reach the ground? (b) What is the velocity of the camcra just before it hits the ground?

A car accelerates uniformly from rest to a speed of \(40.0 \mathrm{mi} / \mathrm{h}\) in \(12.0 \mathrm{~s}\). Find (a) the distance the car travels during this time and (b) the constant acceleration of the car.

Two boats start together and race across a \(60-\mathrm{km}\) -wide lake and back. Boat A goes across at \(60 \mathrm{~km} / \mathrm{h}\) and returns at \(60 \mathrm{~km} / \mathrm{h}\). Boat \(\mathrm{B}\) goes across at \(30 \mathrm{~km} / \mathrm{h}\), and its crew, realizing how far behind it is getting, returns at \(90 \mathrm{~km} / \mathrm{h}\). Turnaround times are negligible, and the boat that completes the round trip first wins. (a) Which boat wins and by how much? (Or is it a tie?) (b) What is the average velocity of the winning boat?

A person sees a lightning bolt pass close to an airplane that is flying in the distance. The person hears thunder \(5.0 \mathrm{~s}\) after seeing the bolt and sees the airplane overhead \(10 \mathrm{~s}\) after hearing the thunder. The speed of sound in air is \(1100 \mathrm{ft} / \mathrm{s}\). (a) Find the distance of the airplane from the person at the instant of the bolt. (Neglect the time it takes the light to travel from the bolt to the eye.) (b) Assuming the plane travels with a constant speed toward the person, find the velocity of the airplanc. (c) Look up the speed of light in air and defend the approximation used in part (a).
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