/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 If a particle is moving in a str... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If a particle is moving in a straight line with constant acceleration and a velocity-time graph is drawn for the motion, the gradient of the graph represents: (a) the acceleration, (b) the rate of increase of velocity, (c) the rate of decrease of velocity.

Short Answer

Expert verified
The gradient represents the acceleration.

Step by step solution

01

Understand the Problem

Determine what the gradient of a velocity-time graph represents when a particle moves with constant acceleration.
02

Define the Gradient

The gradient of a graph is the rate of change of the y-value with respect to the x-value. In a velocity-time graph, the y-value is velocity, and the x-value is time.
03

Meaning of Velocity-Time Graph

On a velocity-time graph, if the acceleration is constant, the graph will be a straight line. The gradient of this straight line is the change in velocity divided by the change in time.
04

Gradient and Acceleration

Since acceleration is defined as the change in velocity over time, it follows that the gradient of the velocity-time graph (change in velocity divided by change in time) represents the acceleration.
05

Conclusion

The gradient of a velocity-time graph for a particle moving with constant acceleration represents the particle's acceleration.

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Key Concepts

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

Constant Acceleration
Constant acceleration means that a particle's velocity changes at a uniform rate over time. If you imagine a car speeding up or slowing down steadily, it’s experiencing constant acceleration. This is a key concept in many physics problems because it simplifies calculations and helps predict future motion.
When a particle has constant acceleration, its velocity-time graph forms a straight line. This straight line indicates a consistent, unchanging rate of speeding up or slowing down. For example, if a car accelerates from 0 to 60 kilometers per hour steadily over 10 seconds, the graph would show a straight line increasing upwards.
Constant acceleration makes it easier to understand physical phenomena and predict trajectories. So, whenever you see a velocity-time graph with a straight line, it often means you're dealing with constant acceleration.
Gradient
The gradient, or slope, is a measure of how steep a line is on a graph. It is crucial in understanding velocity-time graphs. For those graphs, the gradient is calculated as the change in velocity (vertical change) over the change in time (horizontal change). In mathematical terms, it's found using the formula:
\(\text{Gradient} = \frac{ \text{change in velocity} }{ \text{change in time} } \)
A steeper gradient means a higher acceleration and a flatter gradient means a lower acceleration.
Let's break it down:
  • The y-value (vertical axis) represents the velocity.
  • The x-value (horizontal axis) represents the time.
So, a velocity-time graph with a straight line and a steep gradient indicates rapid acceleration, while a more gentle gradient shows slower acceleration. Understanding the gradient helps link the physical world to the graphical representation.
Rate of Change
The rate of change is a broader term that applies to how one quantity changes in relation to another. In a velocity-time graph, this represents how quickly the velocity changes over time.
When dealing with motion, rate of change is synonymous with acceleration. A constant rate of change means constant acceleration. If the rate of change is not constant, the acceleration is varying.
It is important to distinguish between increasing and decreasing rates of change:
  • An increasing rate of change means the velocity is getting bigger over time (speeding up).
  • A decreasing rate of change means the velocity is getting smaller over time (slowing down).
Both situations can occur with constant acceleration, depending on the direction of the acceleration vector. Understanding this helps in predicting how the particle’s motion evolves over time.

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Most popular questions from this chapter

15) Three points A, B, C on a motor racing track are such that \(B\) is \(1 \mathrm{~km}\) beyond A and \(\mathrm{C}\) is \(2 \mathrm{~km}\) beyond B. A car \(\mathrm{X}\), moving with uniform acceleration takes 1 minute to travel from \(A\) to \(B\) and \(1 \frac{1}{2}\) minutes to travel from \(B\) to \(C\). Find its acceleration in \(\mathrm{km} / \mathrm{h} / \mathrm{min}\) and show that its speed at \(\mathrm{C}\) is \(92 \mathrm{~km} / \mathrm{h}\). Another car Y, which is moving with uniform acceleration of \(8 \mathrm{~km} / \mathrm{h}\) per min. passes C 15 . seconds earlier than \(\mathrm{X}\), and its speed is then \(75 \mathrm{~km} / \mathrm{h}\). Find where \(\mathrm{X}\) passes \(\mathrm{Y}\). (Cambridge)

A particle is moving in a straight line. A displacement-time graph is drawn for its motion. The gradient of the tangent to the graph at time \(T\) represents the. velocity of the particle at time \(T\).

A particle is moving in the positive sense on the circumference of a circle of radius \(2 \mathrm{~m}\). The particle has a constant angular acceleration of \(3 \mathrm{rad} \mathrm{s}^{-2}\). At one instant the speed of the particle is \(2 \mathrm{~ms}^{-1}\) and one second later it is \(8 \mathrm{~ms}^{-1}\).

A stone is thrown vertically upward with a speed of \(u\) metre per second. \(A\) second stone is thrown vertically upward from the same point with the same initial speed but \(T\) seconds later than the first one. Prove that they collide at a distance of \(\left(\frac{4 u^{2}-g^{2} T^{2}}{8 g}\right)\) metre above the point of projection.

(a) A particle is moving in a straight line with a constant acceleration of \(2 \mathrm{~ms}^{-2}\) (b) A particle moving in a straight line with a constant acceleration has a velocity of \(2 \mathrm{~ms}^{-1}\) at one instant and a veloctty of \(8 \mathrm{~ms}^{-1}\) three seconds later.

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