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Chapter 3: Problem 6

\(A\) car with a mass of \(1300 \mathrm{~kg}\) is initially moving at a speed of \(40 \mathrm{~km} / \mathrm{h}\) when the brakes are applied and the car is brought to a stop in \(15 \mathrm{~m}\). Assuming that the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

Short Answer

Expert verified
(a) 5330 N, (b) 2.7 s, (c) 4 times, (d) 2 times

Step by step solution

01

Convert Initial Speed

Convert the initial speed from \text{km/h} to \text{m/s}. Given initial speed, \(v = 40 \text{ km/h} = \frac{40 \times 1000}{60 \times 60} \text{ m/s} = 11.1 \text{ m/s}\)
02

Find Acceleration

Use the kinematic equation to find deceleration (negative acceleration). \(\text{Equation: } v^2 = u^2 + 2as\)Where: \(v = 0 \text{ m/s (final speed)}\)\(u = 11.1 \text{ m/s (initial speed)}\)\(s = 15 \text{ m (stopping distance)}\)Solving for \(a\):\(0 = (11.1)^2 + 2 \times a \times 15\)\(a = - \frac{(11.1)^2}{30} = -4.1 \text{ m/s}^2\)
03

Calculate the Force

Use Newton's second law to find the stopping force. \(\text{Force: } F = ma\)Given: \(m = 1300 \text{ kg}\)\(a = -4.1 \text{ m/s}^2\)\(F = 1300 \times -4.1 = -5330 \text{ N}\)The magnitude of the force is \(5330 \text{ N}\)
04

Find the Stopping Time

Use the formula linking acceleration and time for constant deceleration. \(v = u + at\)Solving for \(t\): \(0 = 11.1 + (-4.1)t\)\(t = \frac{11.1}{4.1} = 2.7 \text{ s}\)
05

Stopping Distance with Doubled Initial Speed

Double the initial speed and recalculate stopping distance. New speed: \(u' = 22.2 \text{ m/s}\)Using \(v^2 = u^2 + 2as\): \(0 = (22.2)^2 + 2 \times (-4.1) \times s'\)\(s' = \frac{(22.2)^2}{2 \times 4.1} = 60 \text{ m}\)\(s'\) is 4 times the original stopping distance.
06

Stopping Time with Doubled Initial Speed

Double the initial speed and recalculate stopping time. Using \(v = u + at\): \(0 = 22.2 + (-4.1)t'\)\(t' = \frac{22.2}{4.1} = 5.4 \text{ s}\)\(t'\) is 2 times the original stopping time.

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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 that cause it. It deals with quantities such as displacement, velocity, and acceleration. In this exercise, we use kinematic equations to determine the deceleration and stopping time of a car. The key equation is:

\( v^2 = u^2 + 2as \)

where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration (or deceleration in this case), and \( s \) is the stopping distance. By rearranging this equation, we can solve for the unknown variables.
Newton's Second Law
Newton's second law states that the force acting on an object is equal to the mass of the object times its acceleration:

\( F = ma \)

In this exercise, we use Newton's second law to calculate the force needed to stop the car. The mass of the car \((1300 \text{ kg})\) and the deceleration \((-4.1 \text{ m/s}^2)\) are known, so we can find the required force:

\( F = 1300 \times -4.1 = -5330 \text{ N} \)

The negative sign indicates the direction of the force, which is opposite to the motion.
Deceleration
Deceleration is the rate at which an object slows down. It is the negative of acceleration. In this problem, we find the deceleration using the kinematic equation and the given data. The deceleration is found to be:

\( a = - \frac{(11.1)^2}{30} = -4.1 \text{ m/s}^2 \)

This value tells us how quickly the car slows down over the stopping distance.
Stopping Distance
The stopping distance is the distance a vehicle travels while slowing down to a stop. In the given exercise, we find the stopping distance by rearranging the kinematic equation to solve for \( s \). When the initial speed is doubled, the new stopping distance is calculated as:

\( s' = \frac{(22.2)^2}{2 \times 4.1} = 60 \text{ m} \)

This shows that doubling the initial speed results in a stopping distance that is four times greater.
Initial Speed
Initial speed refers to the speed at which an object is moving when we start measuring its motion. In this problem, the initial speed is given as 40 km/h, which we convert to meters per second:

\( v = 40 \text{ km/h} = \frac{40 \times 1000}{60 \times 60} \text{ m/s} = 11.1 \text{ m/s} \)

Doubling the initial speed to 22.2 m/s impacts both the stopping distance and stopping time, illustrating the importance of initial conditions in motion analysis.

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