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In the 1960 s there was a contest to find the car that could do the following two maneuvers (one right after the other) in the shortest total time: First, accelerate from rest to \(100 \mathrm{mi} / \mathrm{h}(45.0 \mathrm{~m} / \mathrm{s}),\) and then brake to a complete stop. (Ignore the reaction time correction that occurs between the speeding-up and slowing-down phases and assume that all accelerations are constant.) For several years, the winner was the "James Bond car," the Aston Martin. One year it won the contest when it took a total of only 15.0 seconds to perform these two tasks! Its braking acceleration (deceleration) was known to be an excellent \(9.00 \mathrm{~m} / \mathrm{s}^{2}\). (a) Calculate the time it took during the braking phase. (b) Calculate the distance it traveled during the braking phase. (c) Calculate the car's acceleration during the speeding-up phase. (d) Calculate the distance it took to reach \(100 \mathrm{mi} / \mathrm{h}\).

Short Answer

Expert verified
(a) 5.0 s (b) 112.5 m (c) 4.5 m/s² (d) 225 m.

Step by step solution

01

Time for Braking Phase Calculation

To calculate the time it took during the braking phase, we use the formula for uniform acceleration:\[ v = u + at \]where:- \(v\) is the final velocity (0 m/s, since the car stops)- \(u\) is the initial velocity (45.0 m/s)- \(a\) is the acceleration (-9.00 m/s²)Rearranging the formula to solve for \(t\), we get:\[ t = \frac{v-u}{a} = \frac{0 - 45.0}{-9.00} = 5.0 \text{ s} \]
02

Distance During Braking Phase Calculation

To find the distance traveled during the braking phase, we use the equation:\[ s = ut + \frac{1}{2}at^2 \]With \(u = 45.0\,\text{m/s}\), \(a = -9.00\,\text{m/s}^2\), and \(t = 5.0\,\text{s}\), calculate:\[ s = 45.0 \times 5.0 + \frac{1}{2} \times (-9.00) \times (5.0)^2 = 112.5\,\text{m} \]
03

Time for Speeding-Up Phase Calculation

The total time for the entire maneuver was 15.0 s. The time for the braking phase was 5.0 s. Therefore, the time for the speeding-up phase is:\[ t_{\text{acceleration}} = 15.0 - 5.0 = 10.0 \text{ s} \]
04

Acceleration During Speeding-Up Phase Calculation

Use the acceleration formula to find the acceleration during the speeding-up phase:\[ v = u + at \]where \(v = 45.0\,\text{m/s}\), \(u = 0\), \(t = 10.0\,\text{s}\).Rearranging, we get:\[ a = \frac{v-u}{t} = \frac{45.0}{10.0} = 4.5\,\text{m/s}^2 \]
05

Distance During Speeding-Up Phase Calculation

Use the distance formula for the speeding-up phase:\[ s = ut + \frac{1}{2}at^2 \]With \(u = 0\), \(a = 4.5\,\text{m/s}^2\), \(t = 10.0\,\text{s}\), calculate:\[ s = 0 \times 10.0 + \frac{1}{2} \times 4.5 \times (10.0)^2 = 225\,\text{m} \]

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Key Concepts

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

Understanding Kinematics in Motion
When studying motion, kinematics is a fundamental concept that helps us comprehend how objects move. It deals with the mathematical description of motion without considering the forces causing the motion. In kinematics, we often use variables such as displacement, velocity, acceleration, and time to analyze movement. For instance, in the contest to find the fastest car to complete certain maneuvers, understanding kinematics allows us to calculate how long each phase of the car's motion takes and how far it travels in each phase. By closely observing these factors, we can understand how quickly an object like a car can change its state from rest to a given velocity, and then come back to a stop.

In our exercises, kinematics helps us solve different parts of the problem, like finding out the time taken during each phase of motion and computing distances traveled during acceleration and deceleration. Understanding these concepts is invaluable since it builds the foundation for further branches of physics, such as dynamics, which involves the study of forces along with motion.
Exploring Velocity and its Implications
Velocity is a key aspect of kinematics, representing the rate of change of an object's position with respect to time. It is a vector quantity, meaning it has both magnitude and direction. Consider the car from our exercise, which needed to accelerate from rest to 45.0 m/s (equivalent to 100 mi/h) and then brake to a halt. The initial velocity during the acceleration phase is 0 m/s, since the car is at rest. The final velocity is recorded as 45.0 m/s post acceleration.

During the braking phase, the velocity transitions from 45.0 m/s back down to 0 m/s as the car stops. This change in velocity helps us measure the time involved and calculate the rate of acceleration or deceleration. Understanding velocity not only gives insights into how fast an object is moving, but also how its speed can dynamically change during different phases of motion. Being able to calculate acceleration requires a solid understanding of how velocity varies over time.
Examining Braking Distance and its Calculation
Braking distance is the distance that a vehicle travels while slowing down to a stop from a particular speed. This is a critical measurement used to ensure safety and performance of automobiles. In the exercise, the braking distance informs us how far the car traveled from the moment the brakes were applied until it came to a complete halt

In our example, the car starts braking from a speed of 45.0 m/s, and with a braking acceleration of -9.00 m/s², it travels a distance of 112.5 m during the braking phase. Calculating the braking distance involves knowing the initial velocity, acceleration, and the time over which deceleration occurs. The formula used is:
  • \[ s = ut + \frac{1}{2}at^2 \]
where "s" is the distance, "u" is the initial velocity, "a" is the acceleration—and it's negative in the case of braking, and "t" is the time.

This aspect is crucial in vehicle safety analysis, allowing designers to create safer cars by optimizing braking systems and other related technologies.

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Most popular questions from this chapter

A test rocket containing a probe to determine the composition of the upper atmosphere is fired vertically upward from an initial position at ground level. During the time \(t\) while its fuel lasts, the rocket ascends with a constant upward acceleration of magnitude \(2 g .\) Assume that the rocket travels to a small enough height that the Earth's gravitational force can be considered constant. (a) What are the speed and height, in terms of \(g\) and \(t,\) when the rocket's fuel runs out? (b) What is the maximum height the rocket reaches in terms of \(g\) and \(t ?(\mathrm{c})\) If \(t=30.0 \mathrm{~s},\) calculate the rocket's maximum height.

From street level, Superman spots Lois Lane in troublethe evil villain, Lex Luthor, is dropping her from near the top of the Empire State Building. At that very instant, the Man of Steel starts upward at a constant acceleration to attempt a midair rescue of Lois. Assuming she was dropped from a height of \(300 \mathrm{~m}\) and that Superman can accelerate straight upward at \(15.0 \mathrm{~m} / \mathrm{s}^{2},\) determine (a) how far Lois falls before he catches her, (b) how long Superman takes to reach her, and (c) their speeds at the instant he reaches her. Comment on whether these speeds might be a danger to Lois, who, being a mere mortal, might get hurt running into the impervious Man of Steel if the speeds are too great.

The driver of a pickup truck going \(100 \mathrm{~km} / \mathrm{h}\) applies the brakes, giving the truck a uniform deceleration of \(6.50 \mathrm{~m} / \mathrm{s}^{2}\) while it travels \(20.0 \mathrm{~m}\). (a) What is the speed of the truck in kilometers per hour at the end of this distance? (b) How much time has elapsed?

A hockey puck sliding along the ice to the left hits the boards head-on with a speed of \(35 \mathrm{~m} / \mathrm{s}\). As it reverses direction, it is in contact with the boards for \(0.095 \mathrm{~s}\), before rebounding at a slower speed of \(11 \mathrm{~m} / \mathrm{s}\). Determine the average acceleration the puck experienced while hitting the boards. Typical car accelerations are \(5.0 \mathrm{~m} / \mathrm{s}^{2} .\) Comment on the size of your answer, and why it is so different from this value, especially when the puck speeds are similar to car speeds.

A hospital nurse walks \(25 \mathrm{~m}\) to a patient's room at the end of the hall in 0.50 min. She talks with the patient for 4.0 min, and then walks back to the nursing station at the same rate she came. What was the nurse's average speed?

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