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An automobile travels on a straight road for \(40 \mathrm{~km}\) at \(30 \mathrm{~km} / \mathrm{h}\). It then continues in the same direction for an- other \(40 \mathrm{~km}\) at \(60 \mathrm{~km} / \mathrm{h}\). (a) What is the average velocity of the car during the full \(80 \mathrm{~km}\) trip? (Assume that it moves in the positive \(x\) direction.) (b) What is the average speed? (c) Graph \(x\) versus \(t\) and indicate how the average velocity is found on the graph.

Short Answer

Expert verified
(a) 40 km/h, (b) 40 km/h, (c) A straight line slope showing 40 km/h on the graph.

Step by step solution

01

Calculate Time for Each Segment

First, calculate the time taken to travel each 40 km segment. For the first segment at 30 km/h:\[ t_1 = \frac{40 \, \text{km}}{30 \, \text{km/h}} = \frac{4}{3} \, \text{h}\]For the second segment at 60 km/h:\[ t_2 = \frac{40 \, \text{km}}{60 \, \text{km/h}} = \frac{2}{3} \, \text{h}\]
02

Calculate Total Time for the Trip

Add the time for both segments to find the total travel time:\[ t_{\text{total}} = t_1 + t_2 = \frac{4}{3} \, \text{h} + \frac{2}{3} \, \text{h} = 2 \, \text{h}\]
03

Calculate Average Velocity

Average velocity is defined as the total displacement divided by the total time. Since the total displacement is 80 km in the positive direction, the average velocity is:\[ v_{\text{avg}} = \frac{\text{Total displacement}}{\text{Total time}} = \frac{80 \, \text{km}}{2 \, \text{h}} = 40 \, \text{km/h}\]
04

Calculate Average Speed

Average speed is the total distance traveled divided by the total time. In this case, it's the same as the total displacement since the distance equals the displacement.\[ \text{Average speed} = \frac{80 \, \text{km}}{2 \, \text{h}} = 40 \, \text{km/h}\]
05

Sketching the Graph of x vs t

To graph \(x\) versus \(t\), plot time \(t\) on the x-axis and position \(x\) on the y-axis. The graph will have two segments:- A linear segment from \(t=0\) to \(t=\frac{4}{3}\) hours with a slope of 30 km/h (representing the first 40 km).- A linear segment from \(t=\frac{4}{3}\) to \(t=2\) hours with a slope of 60 km/h (representing the next 40 km).The average velocity can be graphically represented as the overall slope from \(x=0\) to \(x=80\) km, forming a straight line over the total time interval from \(t=0\) to \(t=2\) hours.

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Key Concepts

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

Average Speed
Average speed is a fundamental concept in physics that measures how fast an object travels, regardless of its direction. It is calculated as the total distance traveled divided by the total time taken.

This means you add up all the segments of your journey and divide by the time over which these distances were covered.
  • It is a scalar quantity, which means it doesn't have a direction.
  • Unlike velocity, average speed does not account for the direction of travel.
To find the average speed in our exercise:Given that the total distance is 80 km, and the total time is 2 hours, our average speed is \[ \text{Average speed} = \frac{80 \text{ km}}{2 \text{ h}} = 40 \text{ km/h}\] This calculation makes average speed very useful for planning journeys as it gives a straightforward understanding of the time required to cover a specified route under the same conditions.
Displacement
Displacement refers to the overall change in position of an object, from the starting point to the ending point. It is a vector quantity, which means it takes into account both magnitude and direction.

Unlike distance—which measures the total ground covered by the object—displacement measures the shortest path between two points.
  • In linear motion on a straight path, as in this exercise, displacement is equivalent to the total distance if there’s no change in direction.
  • Because our car travels straight without turning, displacement here is simply 80 km in the positive direction.
In practice, displacement can sometimes be zero even if the distance isn't, for instance, if the object returns to its original starting point. But for this journey, since it is a straight trip in one direction, the displacement equals the distance traveled.
Distance-Time Graph
A distance-time graph visually represents how distance changes over time. For a car traveling at different speeds, the distance-time graph offers a quick visual way to understand motion.

To graph the journey, we put time \(t\) on the x-axis and position or distance \(x\) on the y-axis:
  • The first segment from \(t=0\) to \(t=\frac{4}{3}\) hours will have a slope corresponding to 30 km/h.
  • The second segment from \(t=\frac{4}{3}\) to \(t=2\) hours will have a steeper slope of 60 km/h.
The slope of these segments represents the speed during each time interval. The steepness of the slope shows how fast the distance is being covered.

The average velocity can be visualized as a straight line connecting the start and end of the journey. This line's slope is the average velocity over the entire trip. Graphs like these are helpful to easily see differences in speed and to identify uniform, accelerating, or decelerating motion.
Kinematics Equations
Kinematics is the branch of physics that describes the motion of points, objects, or systems without considering the forces that cause this motion. In kinematics, we often use equations to relate several variables of motion: distance, speed, time, and acceleration.

Common kinematic equations are:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
  • \( v_{\text{avg}} = \frac{u + v}{2} \)
Where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is acceleration, \( t \) is time, and \( s \) is displacement.

In scenarios with constant speed and no acceleration, these equations simplify significantly. For linear motion at constant speed, like in our example, the basic equation \( speed = \frac{distance}{time} \) is often adequate. This simplicity is why constant-speed travel problems are often the first step into learning about motion in physics.

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Most popular questions from this chapter

A steel ball is dropped from a building's roof and passes a window, taking \(0.125 \mathrm{~s}\) to fall from the top to the bottom of the window, a distance of \(1.20 \mathrm{~m}\). It then falls to a sidewalk and bounces back past the window, moving from bottom to top in \(0.125 \mathrm{~s}\). Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is \(2.00 \mathrm{~s}\). How tall is the building?

A stone is dropped into a river from a bridge \(43.9 \mathrm{~m}\) above the water. Another stone is thrown vertically down \(1.00 \mathrm{~s}\) after the first is dropped. The stones strike the water at the same time. (a) What is the initial speed of the second stone? (b) Plot velocity versus time on a graph for each stone, taking zero time as the instant the first stone is released.

A key falls from a bridge that is \(45 \mathrm{~m}\) above the water. It falls directly into a model boat, moving with constant velocity, that is \(12 \mathrm{~m}\) from the point of impact when the key is released. What is the speed of the boat?

In \(1 \mathrm{~km}\) races, runner 1 on track 1 (with time \(2 \mathrm{~min}, 27.95 \mathrm{~s}\) ) appears to be faster than runner 2 on track \(2(2 \mathrm{~min}, 28.15 \mathrm{~s})\). However, length \(L_{2}\) of track 2 might be slightly greater than length \(L_{1}\) of track 1. How large can \(L_{2}-L_{1}\) be for us still to conclude that runner 1 is faster?

The brakes on your car can slow you at a rate of \(5.2 \mathrm{~m} / \mathrm{s}^{2}\). (a) If you are going \(137 \mathrm{~km} / \mathrm{h}\) and suddenly see a state trooper, what is the minimum time in which you can get your car under the \(90 \mathrm{~km} / \mathrm{h}\) speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.) (b) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for such a slowing.

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