Slope of the velocity-time graph gives of a moving body. (A) displacement (B) acceleration (C) initial velocity (D) final velocity

Short Answer

Expert verified
The slope of the velocity-time graph represents (B) acceleration, as it is defined as the rate at which velocity changes with respect to time.

Step by step solution

01

Understanding the velocity-time graph

The velocity-time graph is a graphical representation of the motion of a moving object that shows the object's velocity with respect to time. The graph has time represented on the x-axis while the instantaneous velocity is represented on the y-axis.
02

Calculating the slope of the graph

The slope of the graph is the rate at which one variable is changing with respect to another. In the case of a velocity-time graph, the slope represents the rate at which the velocity is changing with respect to time. Mathematically, the slope of the graph can be represented as: Slope = \(\frac{change\:in\:velocity}{change\:in\:time}\)
03

Relating the slope to the options provided

Now, we need to determine which concept from kinematics (displacement, acceleration, initial velocity, or final velocity) is represented by the slope of the velocity-time graph. Let's examine each option: (A) The slope of the graph cannot represent displacement, as displacement is the change in position, not the rate of change of velocity with respect to time. (B) Acceleration is defined as the rate at which velocity changes with respect to time. Therefore, the slope of the velocity-time graph can be represented as acceleration. Mathematically, acceleration is defined as: Acceleration = \(\frac{change\:in\:velocity}{change\:in\:time}\) (C) Initial velocity is the velocity of the body at the start of the time interval or when time is zero. It is not related to the slope of the graph, which represents the rate of change of velocity with respect to time. (D) Final velocity is the velocity of the body at the end of the time interval or when the motion stops. It also does not represent the slope of the graph, which indicates the rate of change of velocity with respect to time. Based on the above analysis, we can determine that:
04

Conclusion

The slope of the velocity-time graph represents (B) acceleration.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The resultant of two vectors \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) can be smaller than \(\mathrm{A}-\mathrm{B}\) in magnitude (B) can be greater than \(\mathrm{A}+\mathrm{B}\) in magnitude (C) can't be greater than \(\mathrm{A}+\mathrm{B}\) or smaller than \(\mathrm{A}-\mathrm{B}\) in magnitude (D) none of above is true

Ball A is thrown in upward from the top of a tower of height h. At the same time ball B starts to fall from that point. When A comes to the top of the tower, B reaches the ground. Find the time to reach maximum height for \(\mathrm{A}\). (A) \(\sqrt{(\mathrm{h} / \mathrm{g})}\) (B) \(\sqrt{(2 \mathrm{~h} / \mathrm{g})}\) (C) \(\sqrt{(h / 2 g)}\) (D) \(\sqrt{(4 \mathrm{~h} / \mathrm{g})}\)

A particle is moving in a straight line with initial velocity of $200 \mathrm{~ms}^{-1}\( acceleration of the particle is given by \)\mathrm{a}=3 \mathrm{t}^{2}-2 \mathrm{t}$. Find velocity of the particle at 10 second. (A) \(1100 \mathrm{~ms}^{-1}\) (B) \(300 \mathrm{~ms}^{-1}\) (C) \(900 \mathrm{~ms}^{-1}\) (D) \(100 \mathrm{~ms}^{-1}\)

The resultant of two forces of magnitude \(2 \mathrm{~N}\) and \(3 \mathrm{~N}\) can never be. (A) \(4 \mathrm{~N}\) (B) \(1 \mathrm{~N}\) (C) \(2.5 \mathrm{~N}\) (D) \((1 / 2) \mathrm{N}\)

A goods train is moving with constant acceleration. When engine passes through a signal its speed is U. Midpoint of the train passes the signal with speed \(\mathrm{V}\). What will be the speed of the last wagon? (B) \(\left.\sqrt{[}\left(\mathrm{V}^{2}-\mathrm{U}^{2}\right) / 2\right]\) (D) \(\sqrt{[}\left(2 \mathrm{~V}^{2}-\mathrm{U}^{2}\right)\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free