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Ch 02: Motion Along a Straight Line
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 2, Problem 10e

A physics professor leaves her house and walks along the sidewalk toward campus. After 55 min, it starts to rain, and she returns home. Her distance from her house as a function of time is shown in Fig. E2.102.10. At which of the labeled points is her velocity decreasing in magnitude?
Position-time graph showing a curve with labeled points I to V; velocity is zero at point IV.

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1
To determine where the velocity is decreasing in magnitude, we need to analyze the slope of the distance-time graph at each labeled point.
Velocity is the derivative of the position with respect to time, which corresponds to the slope of the tangent to the curve at any point.
A decreasing magnitude of velocity means that the slope of the tangent line is becoming less steep, either positively or negatively.
Examine the graph: from point d to e, the slope of the graph is decreasing, indicating that the velocity is decreasing in magnitude.
Therefore, at point e, the velocity is decreasing in magnitude as the slope of the graph is less steep compared to the previous point d.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Velocity

Velocity is a vector quantity that describes the rate of change of position with respect to time, including both speed and direction. On a position-time graph, velocity is represented by the slope of the curve. A steeper slope indicates a higher velocity, while a flat slope indicates zero velocity.
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Slope of a Graph

The slope of a graph at any point is the derivative of the function at that point, representing the rate of change. In a position-time graph, the slope indicates velocity. A decreasing slope means the velocity is decreasing, while an increasing slope means the velocity is increasing.
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Acceleration

Acceleration is the rate of change of velocity with respect to time. It can be positive (increasing velocity) or negative (decreasing velocity). On a position-time graph, acceleration is inferred from changes in the slope; a decreasing slope indicates negative acceleration, meaning the object is slowing down.
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Related Practice
Textbook Question

The table shows test data for the Bugatti Veyron Super Sport, the fastest street car made. The car is moving in a straight line (the xx-axis).

(a) Sketch a vxv_{x}-tt graph of this car's velocity (in mi/h) as a function of time. Is its acceleration constant?

(b) Calculate the car's average acceleration (in m/s2) between (i) 00 and 2.12.1 s; (ii) 2.12.1 s and 20.020.0 s; (iii) 20.020.0 s and 5353 s. Are these results consistent with your graph in part (a)? (Before you decide to buy this car, it might be helpful to know that only 300300 will be built, it runs out of gas in 1212 minutes at top speed, and it costs more than $1.5\$1.5 million!)

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Textbook Question

A physics professor leaves her house and walks along the sidewalk toward campus. After 55 min, it starts to rain, and she returns home. Her distance from her house as a function of time is shown in Fig. E2.102.10. At which of the labeled points is her velocity constant and positive?

3
views
Textbook Question

A physics professor leaves her house and walks along the sidewalk toward campus. After 55 min, it starts to rain, and she returns home. Her distance from her house as a function of time is shown in Fig. E2.102.10. At which of the labeled points is her velocity zero?

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Textbook Question

A physics professor leaves her house and walks along the sidewalk toward campus. After 55 min, it starts to rain, and she returns home. Her distance from her house as a function of time is shown in Fig. E2.102.10. At which of the labeled points is her velocity constant and negative?

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Textbook Question

A race car starts from rest and travels east along a straight and level track. For the first 5.05.0 s of the car's motion, the eastward component of the car's velocity is given by vx(t)=v_{x}(t)= (0.8600.860 m/s3)t2. What is the acceleration of the car when vx=12.0v_{x}=12.0 m/s?

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Textbook Question

A turtle crawls along a straight line, which we will call the xx-axis with the positive direction to the right. The equation for the turtle's position as a function of time is x(t)=50.0x(t) = 50.0 cm + (2.002.00 cm/s)tt − (0.06250.0625 cm/s2)t2t^2. Find the turtle's initial velocity, initial position, and initial acceleration.

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