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Chapter 2: Problem 73

At the instant the traffic light turns green, an automobile starts with a constant acceleration \(a\) of \(2.2 \mathrm{~m} / \mathrm{s}^{2} .\) At the same instant a truck, traveling with a constant speed of \(9.5 \mathrm{~m} / \mathrm{s}\), overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automobile be traveling at that instant?

Short Answer

Expert verified
The car overtakes the truck 82.06 meters beyond the signal, traveling at 19.0 m/s.

Step by step solution

01

Understand the Problem

We have two scenarios 鈥� an accelerating automobile and a constant-speed truck. We need to find when both will be at the same distance from the starting point, meaning the car will have caught up with the truck.
02

Set Equations for the Two Vehicles

First, we define the equations for both cars. The position of the car as a function of time, with initial speed 0 and acceleration \(a = 2.2 \, \text{m/s}^2\), is \(x_c = \frac{1}{2}at^2\). The position of the truck, which is moving at a constant speed of \(v = 9.5 \, \text{m/s}\), is \(x_t = vt\).
03

Establish the Condition for Overtaking

The car overtakes the truck when both have covered the same distance: \(x_c = x_t\). So, set \(\frac{1}{2}at^2 = vt\).
04

Solve for Time

Solve the equation \(\frac{1}{2}(2.2)t^2 = (9.5)t\) for \(t\). Divide both sides by \(t\) (assuming \(teq0\)): \(1.1t = 9.5\). This simplifies to \(t = \frac{9.5}{1.1} \approx 8.64\, \text{s}\).
05

Calculate the Distance When Overtaking Occurs

Substitute \(t = 8.64\, \text{s}\) into the car's position equation: \(x_c = \frac{1}{2}(2.2)(8.64)^2\). Calculate \(x_c \approx 82.06\, \text{m}\). Thus, the car overtakes the truck approximately 82.06 meters beyond the traffic signal.
06

Determine the Automobile's Speed at That Instant

Use the velocity formula for the car, \(v_c = at\). Substitute the known values: \(v_c = 2.2 \times 8.64\, \text{m/s}\). Calculate \(v_c \approx 19.0 \text{m/s}\).

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Key Concepts

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

Constant Acceleration
In kinematics, constant acceleration indicates that an object's velocity is changing at a steady rate. In our exercise, the automobile accelerates with a constant rate of \(2.2 \ \text{m/s}^2\). This means that every second, the vehicle's speed increases by \(2.2 \ \text{m/s}\). There are several things to keep in mind about constant acceleration:
  • It allows the use of certain simple equations to describe motion, since the acceleration is unchanging over time.
  • An object's position as a function of time can be determined using the equation \(x = \frac{1}{2}at^2\) when starting from rest.
  • The velocity at any given time can be found by \(v = at\).
Understanding constant acceleration is crucial because it forms the basis for predicting where an object will be and how fast it will be going over time.
In real-life scenarios like in our exercise, it helps in coordinating the movement of two or more objects. This predicts when one will overtake another based on their speeds and accelerations.
Relative Motion
Relative motion explores how two objects move in relation to each other. In our scenario, we're interested in when the accelerating automobile will catch up with the constant-speed truck. Here's how we deal with relative motion:
  • Instead of looking at each object's motion independently, we compare their positions and velocities.
  • The time it takes for their positions to become equal allows us to say that the automobile overtakes the truck.
  • While the truck travels at \(9.5 \ \text{m/s}\), the car鈥檚 speed continuously increases until they meet.
By setting up equations that compare their positions over time, we use the concept of relative motion to solve for the point of overtaking. This comparative view reveals insights into how different kinematical parameters鈥攍ike speed and acceleration鈥攊nterplay.
Velocity Calculation
Velocity calculation is a fundamental skill in this kinematics exercise. Calculating both vehicles' positions and velocities helps in concluding the point and speed at which the automobile overtakes the truck.For the accelerating automobile:
  • We initially find the time \(t\) it takes for the car to catch up with the truck: \(t \approx 8.64 \ \text{seconds}\).
  • The velocity of the car at this moment is calculated using \(v = at\), so \(v_c = 2.2 \times 8.64 \approx 19.0 \ \text{m/s}\).
This involves understanding not only when the overtaking occurs but also how fast the automobile is moving at that exact instant. Velocity, with its magnitude and direction, allows us to comprehend the comprehensive nature of motion. It's vital for problem-solving where continuous and variable changes in speed occur, as with our accelerating automobile.

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Most popular questions from this chapter

Shows part of a street where traffic flow is to be controlled to allow a platoon of cars to move smoothly along the street. Suppose that the platoon leaders have just reached intersection 2, where the green appeared when they were distance \(d\) from the intersection. They continue to travel at a certain speed \(v_{p}\) (the speed limit) to reach intersection 3, where the green appears when they are distance \(d\) from it. The intersections are separated by distances \(D_{23}\) and \(D_{12}\). (a) What should be the time delay of the onset of green at intersection 3 relative to that at intersection 2 to keep the platoon moving smoothly? Suppose, instead, that the platoon had been stopped by a red light at intersection \(1 .\) When the green comes on there, the leaders require a certain time \(t_{r}\) to respond to the change and an additional time to accelerate at some rate \(a\) to the cruising speed \(v_{p} .\) (b) If the green at intersection 2 is to appear when the leaders are distance \(d\) from that intersection, how long after the light at intersection 1 turns green should the light at intersection 2 turn green?

From \(t=0\) to \(t=5.00 \mathrm{~min}\), a man stands still, and from \(t=5.00 \mathrm{~min}\) to \(t=10.0 \mathrm{~min}\), he walks briskly in a straight line at a constant speed of \(2.20 \mathrm{~m} / \mathrm{s}\). What are (a) his average velocity \(v_{\text {avg }}\) and (b) his average acceleration \(a_{\text {avg }}\) in the time interval \(2.00\) min to \(8.00 \mathrm{~min} ?\) What are (c) \(v_{\text {avg }}\) and (d) \(a_{\text {avg }}\) in the time interval \(3.00 \mathrm{~min}\) to \(9.00 \mathrm{~min} ?\) (e) Sketch \(x\) versus \(t\) and \(v\) versus \(t\), and indicate how the answers to (a) through (d) can be obtained from the graphs.

Shows the speed \(v\) versus height \(y\) of a ball tossed directly upward, along a \(y\) axis. Distance \(d\) is \(0.40 \mathrm{~m}\).The speed at height \(y_{A}\) is \(v_{A} .\) The speed at height \(y_{B}\) is \(\frac{1}{3} v_{A} .\) What is speed \(v_{A}\) ?

Two subway stops are separated by \(1100 \mathrm{~m}\). If a subway train accelerates at \(+1.2 \mathrm{~m} / \mathrm{s}^{2}\) from rest through the first half of the distance and decelerates at \(-1.2 \mathrm{~m} / \mathrm{s}^{2}\) through the second half, what are (a) its travel time and (b) its maximum speed? (c) Graph \(x, v\), and \(a\) versus \(t\) for the trip.

A particle's acceleration along an \(x\) axis is \(a=5.0 t\), with \(t\) in seconds and \(a\) in meters per second squared. At \(t=2.0 \mathrm{~s}\), its velocity is \(+17 \mathrm{~m} / \mathrm{s}\). What is its velocity at \(t=4.0 \mathrm{~s}\) ?
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