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Chapter 2: Problem 42

A certain cable car in San Francisco can stop in \(10 \mathrm{~s}\) when traveling at maximum speed. On one oceasion, the driver sees a dog a distance \(d \mathrm{~m}\) in front of the car and slams on the brakes instantly. The car reaches the dog \(8.0 \mathrm{~s}\) later, and the dog jumps off the track just in time. If the car travels \(4.0 \mathrm{~m}\) beyond the position of the dog before coming to a stop, how far was the car from the dog? (Hint: You will need three equations.)

Short Answer

Expert verified
The solution requires a level of algebra to solve three simultaneous equations. The equations are based off the principles of uniformly accelerated linear motion. Solving the equations will give you the distance 'd'.

Step by step solution

01

Understand the problem and determine what is given

Firstly, understand that we are dealing with a physics problem related to motion. The givens are that: 1. A car can stop in 10 seconds when traveling at maximum speed. 2. The car reaches a dog in front of it 8 seconds after abruptly braking. 3. The car stops 4 meters past the dog. We need to find the initial distance between the car and the dog.
02

Determine the necessary formulas

The equations needed for this problem will be: 1. The formula for uniform acceleration: \(d = u*t + 1/2*a*t^2\), where \(d\) is the displacement, \(u\) is the initial speed, \(a\) is the acceleration, and \(t\) is the time. 2. The equation for final velocity, \(v = u + a*t\), where \(v\) is the final velocity which is zero in both cases as the car stops both times.
03

Find the deceleration

We know that the car takes 10 seconds to stop when at full speed. So we can set the final speed to 0 (as the car stops), and solve for acceleration (which will be a deceleration as it's slowing down) using \(0 = u + a*10\) => \(a = -u/10\). This will be equation (1).
04

Find the distance to the dog

Now, we use the fact that the car reaches the dog in 8 seconds, and the fact that acceleration is constant and equals the deceleration found earlier (as the driver slams on the brakes immediately), and solve for distance using \(d = u*8 + 1/2*a*8^2\). This will give us the distance to where the dog was initially spotted. This will be equation (2).
05

Find the total distance covered by the car

Knowing that the car travels an extra 4 meters after reaching the dog's position before it comes to a stop, find the total distance covered by the car using the equation from Step 2. So, \(d = u*8 + 1/2*a*8^2 + 4\). This will give us the total distance 'd' that the car was initially from the dog. This will be equation (3).
06

Solve the equations to find the answer

Combine equation (1), (2) and (3) to solve for the distance 'd'. As all these equations have two variables (distance and speed), we can equate equations (2) and (3) first, then substitute the 'u' variable from equation (1) to solve for 'd'.

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Key Concepts

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

Uniform Acceleration
Understanding uniform acceleration is essential when solving physics problems involving motion. Uniform acceleration occurs when an object's velocity changes at a constant rate over time. This means the acceleration does not vary, either increasing or decreasing. It remains steady.
In this scenario, you might think about a cable car moving and stopping in a predictable way, as it applies brakes smoothly until it comes to a stop. The rate at which its velocity decreases is constant, meaning there's a uniform deceleration as well. The formula that helps us understand this concept in detail is the equation of motion:\[d = ut + \frac{1}{2} a t^2\]Here, \(d\) represents the total displacement, \(u\) is the initial velocity, \(t\) is the time taken, and \(a\) represents acceleration. In the case of a cable car stopping, the acceleration becomes negative due to deceleration.
This equation provides a clear framework to compute distances covered under uniformly accelerating (or decelerating) conditions, which is a common aspect of physics problem-solving.
Deceleration
Deceleration is simply negative acceleration, meaning the object is slowing down. In our example, as the cable car applies the brakes, it decelerates until it comes to a complete stop.
When calculating deceleration, one critical step is understanding how to compute it using different known variables like initial speed and time.You can calculate the deceleration (negative acceleration) using the following equation:\[v = u + at\]Since the final velocity is zero (as the car comes to a stop), this simplifies to \(0 = u + (-a) \times 10\) if the car stops in 10 seconds. Here, \(a\) is the deceleration, and hence the equation simplifies to \(a = -\frac{u}{10}\).
This formula offers a straightforward way to calculate the rate at which an object slows down over a given period, a critical aspect of many motion-based physics problems.
Distance Calculation
Distance calculation involves determining how far an object travels under certain conditions. In problems surrounding uniform acceleration or deceleration, precisely calculating the total displacement is crucial.
For the scenario of the cable car and the dog, several distances need summing up to provide clarity on the initial distance from the dog.Firstly, you calculate the distance the car covers while decelerating up to the point of meeting the dog:\[d_1 = u \times 8 + \frac{1}{2} a \times 64\]Next, you account for the additional 4 meters it travels beyond the dog before stopping, resulting in the total distance of:\[d_{total} = d_1 + 4\]By combining our understanding of uniform acceleration, deceleration, and these formulae, solving complex physics problems involving motion becomes simpler. Each equation allows us to find hidden quantities and put together a complete picture of the journey an object undertakes.

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Most popular questions from this chapter

A race car moves such that its position fits the relationship $$x=(5.0 \mathrm{~m} / \mathrm{s}) t+\left(0.75 \mathrm{~m} / \mathrm{s}^{3}\right) t^{4}$$ where \(x\) is measured in meters and \(t\) in seconds. (a) Plot a graph of the car's position versus time. (b) Determine the instantaneous velocity of the car at \(t=4.0 \mathrm{~s}\), using time intervals of \(0.40 \mathrm{~s}, 0.20 \mathrm{~s}\), and \(0.10 \mathrm{~s}\). (c) Compare the average velocity during the first \(4.0 \mathrm{~s}\) with the results of part (b).

A steam catapult launches a jet aircraft from the aircraft carrier John C. Stenmis, giving it a speed of \(175 \mathrm{mi} / \mathrm{h}\) in \(2.50 \mathrm{~s}\). (a) Find the average acceleration of the plane. (b) Assuming the acceleration is constant, find the distance the plane moves.

A certain freely falling object requires \(1.50 \mathrm{~s}\) to travel the last \(80.0 \mathrm{~m}\) before it hits the ground. From what height above the ground did it fall?

Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than \(800 \mathrm{~m} / \mathrm{s}^{2}\) lasting for any length of time will not cause injury, whereas an acceleration greater than \(1000 \mathrm{~m} / \mathrm{s}^{2}\) lasting for at least \(1 \mathrm{~ms}\) will cause injury. Suppose a small child rolls off a bed that is \(0.40 \mathrm{~m}\) above the floor. If the floor is hardwood, the child's head is brought to rest in approximately \(2.0 \mathrm{~mm}\). If the floor is carpeted, this stopping distance is increased Io about \(1.0 \mathrm{~cm}\). Calculate the magnitude and duration of the deceleration in both eases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.

A train is traveling down a straight track at \(20 \mathrm{~m} / \mathrm{s}\) when the engincer applies the brakes, resulting in an acceleration of \(-1.0 \mathrm{~m} / \mathrm{s}^{2}\) as long as the train is in motion. How far does the train move during a \(40-\mathrm{s}\) time interval starting at the instant the brakes are applied?
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