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\(\bullet\) Driving home from school one day, you spot a ball rolling out into the street (Figure \(5-18\) ). You brake for \(1.20 \mathrm{s}\), slowing your \(950-\mathrm{kg}\) car from \(16.0 \mathrm{m} / \mathrm{s}\) to \(9.50 \mathrm{m} / \mathrm{s}\). (a) What was the average force exerted on your car during braking? (b) How far did you travel while braking?

Short Answer

Expert verified
(a) The average force is \(-5,149 \, \text{N}\). (b) The car traveled approximately \(15.3 \, \text{m}\) while braking.

Step by step solution

01

Identify known values

We know the initial velocity, \(v_i = 16.0 \, \text{m/s}\), the final velocity, \(v_f = 9.50 \, \text{m/s}\), the mass of the car, \(m = 950 \, \text{kg}\), and the braking time, \(t = 1.20 \, \text{s}\).
02

Calculate acceleration

Use the formula for acceleration, \(a = \frac{v_f - v_i}{t}\).Substitute in the known values: \[a = \frac{9.50 \, \text{m/s} - 16.0 \, \text{m/s}}{1.20 \, \text{s}} = \frac{-6.50 \, \text{m/s}}{1.20 \, \text{s}} \approx -5.42 \, \text{m/s}^2\]
03

Calculate the average force

Apply Newton's second law, \(F = ma\), to find the force exerted on the car.\[F = 950 \, \text{kg} \times (-5.42 \, \text{m/s}^2) = -5,149 \, \text{N}\]The negative sign indicates the force is in the opposite direction of motion.
04

Calculate the braking distance

Use the kinematic equation \(v_f^2 = v_i^2 + 2a d\) to find the distance \(d\).First, rearrange for \(d\): \[d = \frac{v_f^2 - v_i^2}{2a}\]Substitute the known values: \[d = \frac{(9.50 \, \text{m/s})^2 - (16.0 \, \text{m/s})^2}{2 \times -5.42 \, \text{m/s}^2} = \frac{90.25 - 256}{-10.84} \approx 15.3 \, \text{m}\]

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Key Concepts

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

Kinematic Equations
Kinematic equations are very useful for describing the motion of objects. These equations relate five key variables: initial velocity (\(v_i\)), final velocity (\(v_f\)), acceleration (\(a\)), time (\(t\)), and displacement (\(d\)).
In the context of our exercise, these equations allow us to calculate variables like acceleration and displacement when other parameters are known, which is crucial for understanding braking scenarios.
For instance, when a car brakes, we have the initial velocity, the time taken to come to a stop (or slow down), and the final velocity after braking. Using the kinematic equation for acceleration, \(a = \frac{v_f - v_i}{t}\), we can determine how quickly the vehicle is slowing down.
Similarly, the equation \(v_f^2 = v_i^2 + 2ad\) helps us calculate how far the car traveled during the braking process by solving for displacement (\(d\)).
Understanding how to manipulate these equations opens up a world of insight into the physics of motion.
Average Force Calculation
Calculating the average force exerted during braking is crucial in physics when assessing how external interactions affect motion. Newton's Second Law, \(F = ma\), is central to solving such problems, where \(F\) represents the force, \(m\) the mass of the object, and \(a\) the acceleration.
When a car brakes, an average force is exerted in the opposite direction of movement to decelerate it. In our scenario, the car weighs 950 kg and experiences an acceleration of \(-5.42 \, \text{m/s}^2\), calculated earlier using kinematic equations.
By plugging these values into the formula, the average braking force can be found: \(F = 950 \, \text{kg} \times (-5.42 \, \text{m/s}^2) = -5,149 \, \text{N}\).
The negative sign indicates the force direction is opposite to the car's initial direction of motion. Calculating such forces helps to design safer vehicles and understand vehicular dynamics more comprehensively.
Braking Distance
Understanding braking distance is essential for safety in driving. It's the distance a vehicle travels from the moment the brakes are applied until it comes to a complete stop. This distance can be calculated using the kinematic equation: \(v_f^2 = v_i^2 + 2ad\).
Rearranging this to solve for distance gives us \(d = \frac{v_f^2 - v_i^2}{2a}\), where \(v_f\) and \(v_i\) are the final and initial velocities, respectively, and \(a\) is the acceleration.
In the scenario presented, we found the braking distance by substituting the known values into this formula, resulting in a distance of approximately 15.3 m.
This practical understanding of braking distances helps drivers maintain safe following distances and understand how speed affects stopping capability. It's a core concept in physics that plays a critical role in automotive safety engineering.

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Most popular questions from this chapter

Suppose the initial speed of the driver is doubled to \(36.0 \mathrm{m} / \mathrm{s}\). If the driver still has a mass of \(65.0 \mathrm{kg}\), and comes to rest in \(1.00 \mathrm{m},\) what is the magnitude of the force exerted on the driver during this collision? A. \(648 \mathrm{N}\) B. \(1170 \mathrm{N}\) C. \(2.11 \times 10^{4} \mathrm{N}\) D. \(4.21 \times 10^{4} \mathrm{N}\)

IP An archer shoots a 0.024-kg arrow at a target with a speed of \(54 \mathrm{m} / \mathrm{s}\). When it hits the target, it penetrates to a depth of \(0.083 \mathrm{m} .\) (a) What was the average force exerted by the target on the arrow? (b) If the mass of the arrow is doubled, and the force exerted by the target on the arrow remains the same, by what multiplicative factor does the penetration depth change? Explain.

\- ce Predict/Explain You drop two balls of equal diameter from the same height at the same time. Ball 1 is made of metal and has a greater mass than ball \(2,\) which is made of wood. The upward force due to air resistance is the same for both balls. (a) Is the drop time of ball 1 greater than, less than, or equal to the drop time of ball \(2 ?\) (b) Choose the best explanation from among the following: I. The acceleration of gravity is the same for all objects, regardless of mass. II. The more massive ball is harder to accelerate. III. Air resistance has less effect on the more massive ball.

Your groceries are in a bag with paper handles. The handles will tear off if a force greater than \(51.5 \mathrm{N}\) is applied to them. What is the greatest mass of groceries that can be lifted safely with this bag, given that the bag is raised (a) with constant speed, or (b) with an acceleration of \(1.25 \mathrm{m} / \mathrm{s}^{2} ?\)

In baseball, a pitcher can accelerate a \(0.15-\mathrm{kg}\) ball from rest to \(98 \mathrm{mi} / \mathrm{h}\) in a distance of \(1.7 \mathrm{m}\). (a) What is the average force exerted on the ball during the pitch? (b) If the mass of the ball is increased, is the force required of the pitcher increased, decreased, or unchanged? Explain.

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