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Chapter 2: Problem 40

A car starts from rest and travels for \(t_{1}\) seconds with a uniform acceleration \(a_{1}\). The driver then applies the brakes, causing a uniform acceleration \(a_{2}\). If the brakes are applied for \(t_{2}\) seconds, (a) how fast is the car going just before the beginning of the braking period? (b) How far does the car go before the driver begins to brake? (c) Using the answers to parts (a) and (b) as the initial velocity and position for the motion of the car during braking, what total distance does the car travel? Answers are in terms of the variables \(a_{1}, a_{2}, t_{1}\), and \(t_{2}\).

Short Answer

Expert verified
The speed at the beginning of the braking period is \(a_{1} \cdot t_{1}\), the distance before the driver begins to brake is \(0.5 \cdot a_{1} \cdot t_{1}^2\), and the total distance traveled by the car is \(0.5 \cdot a_{1} \cdot t_{1}^2 - \frac{a_{1} \cdot t_{1}^2}{a_{2}}\)

Step by step solution

01

Calculate the speed at the beginning of the braking period

Apply the first equation of motion, \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration and \(t\) is the time. For the first period, since the car starts from rest, the initial velocity \(u = 0\). So, the final velocity at the beginning of the braking period or the speed \(v_{1}\) before braking starts can be calculated as \(v_{1} = a_{1} \cdot t_{1}\)
02

Calculate the distance before the driver begins to brake

Apply the second equation of motion, \(s = ut + 0.5 \cdot a \cdot t^2\), where \(s\) is the distance covered. As the car starts from rest, \(u = 0\). The distance \(d_{1}\) covered during the first period is therefore \(d_{1} = 0.5 \cdot a_{1} \cdot t_{1}^2\)
03

Calculate the speed after the braking period

Applying again the first equation of motion for the braking period, now taking the initial velocity as \(v_{1}\) (found from Step 1) and acceleration as \(a_{2}\). As the car comes to a halt after braking, the final velocity \(v = 0\). So, the time \(t_{2}\) during braking is calculated as \(t_{2} = -\frac{v_{1}}{a_{2}} = -\frac{a_{1} \cdot t_{1}}{a_{2}}\) (note the negative sign as the acceleration is now in the opposite direction)
04

Calculate the distance covered during braking

Again applying the second equation of motion during the braking period, taking initial velocity as \(v_{1}\), acceleration as \(a_{2}\) and time as \(t_{2}\). The distance \(d_{2}\) covered during braking is calculated as \(d_{2} = v_{1} \cdot t_{2} + 0.5 \cdot a_{2} \cdot t_{2}^2\) which simplifies to \(d_{2} = -\frac{a_{1} \cdot t_{1}^2}{a_{2}}\)
05

Calculate total distance covered

Finally, calculate the total distance \(d = d_{1} + d_{2}\) covered, which is the sum of distances covered during acceleration and braking. Therefore, \(d = 0.5 \cdot a_{1} \cdot t_{1}^2 - \frac{a_{1} \cdot t_{1}^2}{a_{2}}\)

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Key Concepts

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

Uniform Acceleration
In physics, uniform acceleration refers to the consistent increase or decrease in velocity over time. One of the most common examples of uniform acceleration is that of a car that starts from rest and speeds up at a steady rate until the brakes are applied. In this scenario, acceleration is described by a constant, which is denoted as a_1 for the period of speeding up and a_2 for the period of braking.

Equations of Motion
Equations that describe the motion of objects are crucial tools in kinematics. They're known as equations of motion, and they provide the relationship between velocity, acceleration, distance, and time. The most commonly used equations of motion are:
  • v = u + at: This equation gives the final velocity (v) of an object given its initial velocity (u), uniform acceleration (a), and time (t) elapsed.
  • s = ut + 0.5 * a * t^2: It calculates the distance (s) traveled by an object, again based on its initial velocity (u), acceleration (a), and time (t) of travel.
These equations apply to uniformly accelerated motion and are remarkably handy for solving problems like the movement of a car braking and accelerating.

Kinematic Distance Calculation
Calculating the distance covered by an object under uniform acceleration involves using kinematic formulas. In the case of the exercise, we calculated the distance during both acceleration and deceleration (braking). For kinematic distance calculation, the formula s = ut + 0.5 * a * t^2 was used to find the distance before braking (d_1). During the braking phase, even though the velocity is decreasing, the same formula can be applied because acceleration (albeit negative, as it's deceleration) remains consistent. By adding the distances covered during both the acceleration and braking phases (d_1 + d_2), we can determine the total distance traveled by the car.

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Most popular questions from this chapter

Two students are on a balcony a distance \(h\) above the street. One student throws a ball vertically downward at a speed \(v_{0}\); at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of \(v_{0}, g_{1} h\), and \(\ell\). (a) Write the kinematic equation for the \(y\) -coordinate of each ball. (b) Set the equations found in part (a) equal to height 0 and solve each for \(t\) symbolically using the quadratic formula. What is the difference in the two balls' time in the air? (c) Use the time-independent kinematics equation to find the velocity of each ball as it strikes the ground. (d) How far apart are the balls at a time \(t\) after they are released and before they strike the ground?

Two students are on a balcony \(19.6 \mathrm{~m}\) above the street. One student throws a ball vertically downward at \(14.7 \mathrm{~m} / \mathrm{s} ;\) at the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. (a) What is the difference in the two balls' time in the air? (b) What is the velocity of each ball as it strikes the ground? (c) How far apart are the balls \(0.800 \mathrm{~s}\) after they are thrown?

The average person passes out at an acceleration of \(7 \mathrm{~g}\) (that is, seven times the gravitational acceleration on Earth). Suppose a car is designed to accelerate at this rate. How much time would be required for the car to accelerate from rest to \(60.0\) miles per hour? (The car would need rocket boosters!)

A record of travel along a straight path is as follows: 1\. Start from rest with a constant acceleration of \(2.77 \mathrm{~m} / \mathrm{s}^{2}\) for \(15.0 \mathrm{~s}\) 2\. Maintain a constant velocity for the nexL \(2.05 \mathrm{~min}\). 3\. Apply a constant negative acceleration of \(-9.47 \mathrm{~m} / \mathrm{s}^{2}\) for \(4.39 \mathrm{~s}\). (a) What was the total displacement for the trip? (b) What were the average speeds for legs 1,2, and 3 of the trip, as well as for the complete trip?

A train \(400 \mathrm{~m}\) long is moving on a straight track with a speed of \(82.4 \mathrm{~km} / \mathrm{h}\). The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of \(16.4 \mathrm{~km} / \mathrm{h}\). Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.
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