/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Coasting due west on your bicycl... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 60

Coasting due west on your bicycle at \(8.4 \mathrm{m} / \mathrm{s}\), you encounter a sandy patch of road \(7.2 \mathrm{m}\) across. When you leave the sandy patch your speed has been reduced by \(2.0 \mathrm{m} / \mathrm{s}\) to \(6.4 \mathrm{m} / \mathrm{s} .\) (a) Assuming the sand causes a constant acceleration, what was the bicycle's acceleration in the sandy patch? Give both magnitude and direction. (b) How long did it take to cross the sandy patch? (c) Suppose you enter the sandy patch with a speed of only \(5.4 \mathrm{m} / \mathrm{s} .\) Is your final speed in this case \(3.4 \mathrm{m} / \mathrm{s}\) more than \(3.4 \mathrm{m} / \mathrm{s},\) or less than \(3.4 \mathrm{m} / \mathrm{s}\) ? Explain.

Short Answer

Expert verified
(a) Acceleration is \(-2.06 \, \text{m/s}^2\). (b) Time is \(0.97\, \text{s}\). (c) Final speed is less than \(3.4 \, \text{m/s}\).

Step by step solution

01

Initial Assumptions

The cyclist encounters a sandy patch of length \(7.2\, \text{m}\), and the speed reduces from \(8.4\, \text{m/s}\) to \(6.4\, \text{m/s}\). This reduction occurs due to constant acceleration over the sandy patch.
02

Finding Acceleration (a)

We use the equation of motion: \[v^2 = u^2 + 2as \]where: \( v = 6.4 \, \text{m/s} \) (final speed), \( u = 8.4 \, \text{m/s} \) (initial speed), \( s = 7.2 \, \text{m} \) (distance).Rearrange the equation for \( a \):\[a = \frac{v^2 - u^2}{2s} \]Substitute the known values:\[a = \frac{6.4^2 - 8.4^2}{2 \times 7.2} \]\[a = \frac{40.96 - 70.56}{14.4}\]\[a = \frac{-29.6}{14.4} \]\[a = -2.06 \, \text{m/s}^2\]The negative sign indicates the acceleration is opposite to the direction of motion (west).
03

Calculating the Time Interval (b)

Using the equation:\[v = u + at\]Solve for \( t \): \[t = \frac{v - u}{a}\]Substitute the known values, including \( a = -2.06 \, \text{m/s}^2 \):\[t = \frac{6.4 - 8.4}{-2.06} \]\[t = \frac{-2.0}{-2.06} \]\[t \approx 0.97 \, \text{s}\]
04

Final Speed with Reduced Initial Speed (c)

To check if the final speed is more or less than \(3.4\,\text{m/s}\), use the same equation of motion as in part (a):\[v^2 = u^2 + 2as\]Set \( u = 5.4 \, \text{m/s} \) and use the same acceleration and distance.Calculate:\[v^2 = 5.4^2 + 2(-2.06)(7.2)\]\[v^2 = 29.16 - 29.65\]\[v^2 = -0.49\]Since \( v^2 \) is negative, the motion doesn't physically end with a speed valid by real numbers, indicating it would stop before exiting the patch, implying final speed is \(0\,\text{m/s}\), which is less than \(3.4\,\text{m/s}\).

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Key Concepts

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

Constant Acceleration
In kinematics, acceleration refers to the rate of change of velocity over time. When an object, such as a bicycle, moves through a medium with a constant acceleration, it experiences a steady increase or decrease in its speed. This concept was central to solving the exercise where the cyclist's speed changed as they crossed a sandy patch. Constant acceleration can be either positive or negative but remains uniform across the distance. In this exercise, the sand causes negative constant acceleration, meaning it slows the cyclist down uniformly as they pass through it. The calculation involves understanding that the acceleration due to the sand is consistent across the entire sandy stretch, leading to a predictable change in speed. This is why constant acceleration equations are so reliable鈥攖hey allow us to predict exactly how speed will change over time or distance.
Equations of Motion
The equations of motion are a set of equations used to predict the future location and velocity of a moving object under the effect of constant acceleration. These are essential tools in physics for solving mechanics problems. In the exercise, the equation \[v^2 = u^2 + 2as\] was used to find the acceleration. Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance traveled. This equation is one of the three core equations that address different scenarios of motion, depending on what is known and what needs to be calculated.Some key points about the equations include:
  • They apply only under constant acceleration conditions.
  • Each equation provides insights into motion aspects like position, speed, or time.
  • Rearranging these equations can solve for unknown variables.
These equations help us understand how different factors like speed, distance, and time interrelate when acceleration does not change.
Problem Solving in Physics
When tackling physics problems, especially in kinematics, a structured approach is beneficial. The first step is to list known quantities and identify what needs to be solved. For example, when the cyclist enters a sandy patch, we know the initial speed, final speed, and distance traveled, and need to find acceleration and time. Here鈥檚 how to approach such a problem:
  • Identify: What is given, and what you need to find.
  • Choose: The appropriate equations of motion based on known and unknown variables.
  • Solve: Substitute known values into the equation, and solve for the unknown.
  • Check: Run through calculations to ensure they make physical sense.
Adopting a methodical approach improves problem-solving skills and comprehension. Mistakes become learning opportunities, revealing deeper insights into physical principles. This way, complex problems become manageable and more intuitive.

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Most popular questions from this chapter

A car and a truck are heading directly toward one another on a straight and narrow street, but they avoid a head-on collision by simultaneously applying their brakes at \(t=0\). The resulting velocity-versus-time graphs are shown in Figure \(2-31\). What is the separation between the car and the truck when they have come to rest, given that at \(t=0\) the car is at \(x=15 \mathrm{m}\) and the truck is at \(x=-35 \mathrm{m} ?\) (Note that this information determines which line in the graph corresponds to which vehicle.)

Suppose the balloon is descending with a constant speed of \(4.2 \mathrm{m} / \mathrm{s}\) when the bag of sand comes loose at a height of \(35 \mathrm{m}\). (a) How long is the bag in the air? (b) What is the speed of the bag when it is \(15 \mathrm{m}\) above the ground?

You drop a ski glove from a height \(h\) onto fresh snow, and it sinks to a depth \(d\) before coming to rest. (a) In terms of \(g\) and \(h,\) what is the speed of the glove when it reaches the snow? (b) What are the magnitude and direction of the glove's acceleration as it moves through the snow, assuming it to be constant? Give your answer in terms of \(g, h,\) and \(d\).

The Detonator On a ride called the Detonator at Worlds of Fun in Kansas City, passengers accelerate straight downward from rest to \(45 \mathrm{mi} / \mathrm{h}\) in \(2.2 \mathrm{seconds.}\) What is the average acceleration of the passengers on this ride?

A popular entertainment at some carnivals is the blanket toss (see photo, p. 39 ). (a) If a person is thrown to a maximum height of \(28.0 \mathrm{ft}\) above the blanket, how long does she spend in the air? (b) Is the amount of time the person is above a height of \(14.0 \mathrm{ft}\) more than, less than, or equal to the amount of time the person is below a height of \(14.0 \mathrm{ft}\) ? Explain. (c) Verify your answer to part (b) with a calculation.
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