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Chapter 2: Problem 72

Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up for \(402 \mathrm{~m}\) with an acceleration of \(+17.0 \mathrm{~m} / \mathrm{s}^{2}\). A parachute then opens, slowing the car down with an acceleration of \(-6.10 \mathrm{~m} / \mathrm{s}^{2} .\) How fast is the racer moving \(3.50 \times 10^{2} \mathrm{~m}\) after the parachute opens?

Short Answer

Expert verified
The racer's speed is approximately 97.1 m/s after 350 m from the parachute opening.

Step by step solution

01

Calculate Final Velocity After Initial Acceleration

The first step is to determine the velocity of the drag racer after it has accelerated from rest for 402 m. We can use the kinematic equation:\[v_f^2 = v_i^2 + 2a_1d\]where \(v_f\) is the final velocity, \(v_i = 0\ \mathrm{m/s}\) (initial velocity), \(a_1 = 17.0\ \mathrm{m/s}^2\) (acceleration), and \(d = 402\ \mathrm{m}\) (distance). Substituting the values:\[v_f^2 = 0 + 2 \times 17.0 \times 402 = 13668\]Solving for \(v_f\):\[v_f = \sqrt{13668} \approx 117.0\ \mathrm{m/s}\]
02

Define the Initial Conditions for Deceleration

After reaching a velocity of 117.0 m/s, the parachute opens, and the racer begins to decelerate. The initial velocity for the deceleration phase is then \(v_i = 117\ \mathrm{m/s}\), and the distance that we need to consider is 350 m.
03

Calculate Final Velocity After Deceleration

To find the velocity after deceleration over 350 m, we use the kinematic equation again:\[v_f^2 = v_i^2 + 2a_2d\]where \(a_2 = -6.10\ \mathrm{m/s}^2\) (deceleration) and \(d = 350\ \mathrm{m}\). Substituting the known values:\[v_f^2 = 117^2 + 2 \times (-6.10) \times 350\]Simplify and calculate:\[v_f^2 = 13689 - 4270 = 9419\]Then solving for \(v_f\):\[v_f = \sqrt{9419} \approx 97.1\ \mathrm{m/s}\]

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Key Concepts

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

Understanding Acceleration
Acceleration is the rate at which an object changes its velocity over time. It's a vector quantity, which means both its magnitude and direction matter. In the context of the drag racer, the acceleration was
  • Positive, at +17.0 m/s虏, indicating a speeding up.
  • Applied over a distance of 402 meters.
This means the car's velocity increased steadily as it moved, starting from a complete stop. When dealing with acceleration, the kinematic equation \[v_f^2 = v_i^2 + 2ad\]helps us find the final velocity after a specific distance. Here, since the initial velocity \(v_i\) was 0, the racer accelerated to approximately 117 m/s within 402 meters.

Acceleration is crucial in determining how fast an object will be over a given distance and time. It's influenced by the forces applied and the mass of the object.
Deceleration Explained
Deceleration is simply acceleration in the opposite direction of the motion. It's commonly thought of as slowing down, and it occurs when a force, such as a parachute, is applied. For our drag racer, the deceleration value is
  • -6.10 m/s虏, indicating a reduction in speed.
When the parachute opens, it creates a backward force, causing the drag racer to slow down over a distance of 350 meters.

We use the same kinematic equation for deceleration:\[v_f^2 = v_i^2 + 2ad\]By plugging in the values (with a negative acceleration due to the opposing force), we can find the final velocity as the racer slows, which ends up being approximately 97.1 m/s. Understanding deceleration helps us predict how quickly an object will come to a stop or reach a reduced speed after opposing forces act on it.
Kinematic Equations in Action
Kinematic equations are essential in solving motion-related problems, especially those involving constant acceleration or deceleration. These equations allow us to calculate unknown variables such as distance, velocity, or acceleration when others are known.

In this exercise, the key kinematic equation used was:
  • \[v_f^2 = v_i^2 + 2ad\]
The equation was used twice. First, to find the final velocity after initial acceleration, and second, to determine the new final velocity after deceleration. These formulas simplify the process of analyzing motion, removing the need for complicated differential equations. They provide clear insight into the effects of forces like speeding up and slowing down, making them indispensable in physics.
Essentials of Velocity Calculations
Velocity calculations help in determining how fast an object is moving in a specific direction. In both scenarios of this exercise鈥攖he initial acceleration and subsequent deceleration鈥攖he velocity was calculated by considering:
  • The initial velocity at the start of each phase.
  • The acceleration or deceleration.
  • The distance covered.
First, starting from rest, the drag racer reached 117 m/s after accelerating for 402 meters. Then, while decelerating with the parachute over 350 meters, it slowed to about 97.1 m/s.

Velocity is different from speed as it accounts for direction and is a key parameter when assessing motion. Calculating velocity correctly ensures accurate interpretations of how an object鈥檚 speed and direction change over time.

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Most popular questions from this chapter

A woman and her dog are out for a morning run to the river, which is located \(4.0 \mathrm{~km}\) away. The woman runs at \(2.5 \mathrm{~m} / \mathrm{s}\) in a straight line. The dog is unleashed and runs back and forth at \(4.5 \mathrm{~m} / \mathrm{s}\) between his owner and the river, until the woman reaches the river. What is the total distance run by the dog?

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 22 minutes at an average speed of \(7.2 \mathrm{~m} / \mathrm{s}\). During the second part, she rides for 36 minutes at an average speed of \(5.1 \mathrm{~m} / \mathrm{s} .\) Finally, during the third part, she rides for 8.0 minutes at an average speed of 13 \(\mathrm{m} / \mathrm{s}\). (a) How far has the bicyclist traveled during the entire trip? (b) What is her average velocity for the trip?

Review Interactive LearningWare 2.2 at in preparation for this problem. A race driver has made a pit stop to refuel. After refueling, he leaves the pit area with an acceleration whose magnitude is \(6.0 \mathrm{~m} / \mathrm{s}^{2}\); after \(4.0 \mathrm{~s}\) he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant speed of \(70.0 \mathrm{~m} / \mathrm{s}\) overtakes and passes the entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?

The initial velocity \(v_{0}\) and acceleration \(a\) of four moving objects at a given instant in time are given in the table: $$ \begin{array}{|c|c|c|} \hline & \text { Initial velocity } v_{0} & \text { Acceleration } a \\ \hline \text { (a) } & +12 \mathrm{~m} / \mathrm{s} & +3.0 \mathrm{~m} / \mathrm{s}^{2} \\ \hline \text { (b) } & +12 \mathrm{~m} / \mathrm{s} & -3.0 \mathrm{~m} / \mathrm{s}^{2} \\ \hline \text { (c) } & -12 \mathrm{~m} / \mathrm{s} & +3.0 \mathrm{~m} / \mathrm{s}^{2} \\ \hline \text { (d) } & -12 \mathrm{~m} / \mathrm{s} & -3.0 \mathrm{~m} / \mathrm{s}^{2} \\ \hline \end{array} $$ Draw vectors for \(v_{0}\) and \(a\) and, in each case, state whether the speed of the object is increasing or decreasing in time. Account for your answers. Problem For each of the four pairs in the table above, determine the final speed of the object if the elapsed time is \(2.0 \mathrm{~s}\). Compare your final speeds with the initial speeds and make sure that your answers are consistent with your answers to the Concept Questions.

An 18 -year-old runner can complete a \(10.0-\mathrm{km}\) course with an average speed of \(4.39 \mathrm{~m} / \mathrm{s}\). A 50-year-old runner can cover the same distance with an average speed of \(4.27 \mathrm{~m} / \mathrm{s}\). How much later (in seconds) should the younger runner start in order to finish the course at the same time as the older runner?
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