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Acceleration and velocity-time graphs

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Summary

Acceleration and velocity-time graphs

In a nutshell

Acceleration is a vector, and is the rate of change of velocity. Acceleration over a period of time can be calculated by determining the gradient of a velocity time graph.

Equations

DESCRIPTION	EQUATION
Velocity	$v = \dfrac{\Delta s}{\Delta t}$
Acceleration	$a = \dfrac{\Delta v}{\Delta t}$

Variables

QUANTITY NAME	SYMBOL	DERIVED UNIT	SI UNIT
$speed$	$v$	$ms^{-1}$	$ms^{-1}$
$velocity$	$v$	$ms^{-1}$	$ms^{-1}$
$displacement$	$s$	$m$	$m$
$time$	$t$	$s$	$s$
$acceleration$	$a$	$ms^{-2}$	$ms^{-2}$

Acceleration

Acceleration is defined as the rate of change of velocity. It is a vector, as it has both a magnitude and a direction. A negative acceleration is also sometimes referred to as a deceleration. It can be calculated using the following equation:

$a = \dfrac{\Delta v}{\Delta t}$

Note: due to this definition, an object can accelerate by changing its direction or speed, or both.

Velocity-time graphs

The motion of an object can also be expressed on a velocity-time graph. On this graph, the y-coordinate represents the instantaneous velocity of the object. The gradient of the graph represents the acceleration. If it is a straight line acceleration is constant, if it is curved the acceleration is changing.

The area under the graph is the displacement travelled. For a straight-line graph, this is easy to calculate, as the area under the graph is a series of triangles and rectangles which you can just add the area of together.

Physics; Motion; KS5 Year 12; Acceleration and velocity-time graphs

1.	In this section, the shape is that of a rectangle, so you would need to multiply the base and the height of the shape to find the area.
2.	In this section, the shape is that of a triangle, so you would need to multiply the base and the height together and divide by 2.

Example:

The velocity-time graph of a smart car is shown in the figure below. Calculate the total distance travelled by the car in the $50\ s$ period.

First state the equation:

$distance \ travelled= area \ under \ the \ graph$

Then calculate the answer:

$0<t<10 \ s: 0.5 \times 60 \times 10 = 300\ m \\[0.05in] \newline 10<t<40 \ s: 60 \times 30 = 1,800 \ m \\[0.05in] \newline 40<t<50 \ s:0.5 \times 60 \times 10 = 300\ m\newline \\[0.1in] area \ under = 300 + 1,800 + 300\newline = 2,400\ m$

The distance travelled by the smart car in the $50 \ s$ period is $\underline{2400\ m}$ .

Calculating displacement for changing accelerations

As described before, when the acceleration is changing, the velocity-time graph will show a curved line. This means that the area under the curve has to be estimated by drawing a series of shapes with calculatable areas (triangles, rectangles and trapeziums). The total area will be the sum of the individual shapes areas.

To get the instantaneous velocity, draw a tangent to the curve at the specified time and calculate the gradient of this tangent.

Learn with Basics

Length:

Unit 1

Motion and speed

Unit 2

Speed, velocity and acceleration

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Unit 3