How To Draw A Acceleration Vs Time Graph From A Velocity Time Graph

It was learned earlier in Lesson 4 that the slope of the line on a velocity versus time graph is equal to the acceleration of the object. If the object is moving with an acceleration of +4 m/southward/due south (i.eastward., changing its velocity by iv m/s per second), so the slope of the line will be +4 k/s/s. If the object is moving with an acceleration of -eight m/southward/s, so the slope of the line will be -viii 1000/south/s. If the object has a velocity of 0 m/southward, so the slope of the line will exist 0 m/southward. Because of its importance, a student of physics must take a expert understanding of how to calculate the slope of a line. In this part of the lesson, the method for determining the slope of a line on a velocity-time graph will exist discussed.

Permit's begin past considering the velocity versus time graph below.

The line is sloping upwards to the correct. But mathematically, by how much does it slope upwards for every 1 2nd along the horizontal (fourth dimension) axis? To answer this question nosotros must use the gradient equation.

Using the Gradient Equation

The slope equation says that the slope of a line is institute past determining the amount of ascension of the line betwixt whatsoever two points divided by the amount of run of the line between the aforementioned two points. A method for carrying out the adding is

Pick ii points on the line and determine their coordinates.
Determine the difference in y-coordinates for these two points (rise).
Determine the divergence in ten-coordinates for these two points (run).
Split the difference in y-coordinates by the deviation in 10-coordinates (rise/run or slope).

The calculations below shows how this method can be applied to determine the gradient of the line. Notation that three different calculations are performed for iii different sets of two points on the line. In each case, the effect is the same: the slope is 10 m/s/s.

For points (v s, 50 chiliad/s) and (0 south, 0 m/southward):

Slope = (50 k/s - 0 m/southward) / (5 s - 0 s) = x m/s/s

For points (5 s, 50 thou/southward) and (two s, twenty m/southward):

Slope = (fifty m/south - 20 chiliad/south) / (5 south - 2 southward) = x m/southward/s

For points (iv s, forty m/s) and (3 s, xxx grand/south):

Gradient = (40 chiliad/s - 30 yard/southward) / (4 s - 3 s) = 10 g/s/s

Find that regardless of which two points on the line are chosen for the slope calculation, the result remains the aforementioned - 10 m/s/s.

Check Your Understanding

Consider the velocity-time graph below. Determine the acceleration (i.due east., slope) of the object every bit portrayed by the graph. Use the button to view the answer.

We Would Like to Propose ...

Sometimes information technology isn't enough to just read about it. You have to collaborate with it! And that's exactly what y'all do when you lot use ane of The Physics Classroom's Interactives. Nosotros would like to suggest that you combine the reading of this page with the use of our Two Phase Rocket Interactive. This Interactive is found in the Physics Interactives section of our website and allows a learner to apply the skill of calculating slopes and relating them to acceleration values for a two-stage rocket.