The Acceleration-Time graph speak us about an object’s velocity the same means the Velocity Time graph tells us around an object’s displacement. In this article, you will learn about acceleration graphs in detail.

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## What is Acceleration-Time Graph?

Acceleration-Time Graph mirrors the acceleration plotted against time because that a particle relocating in a straight line. The acceleration-time plots acceleration values on the y-axis and time worths on the x-axis.

### Vertical Axis

The upright axis in the acceleration-time graph to represent the acceleration the the object. In the offered graph below, reading the worth of the graph in ~ a certain time will certainly fetch girlfriend the acceleration of the thing in meters per 2nd squared for that moment.

Acceleration-Time Graph

### Slope of the acceleration graph

The slope of the acceleration graph represents a quantity known as a jerk. Jerk is the rate of change of acceleration. In the offered acceleration graph as displayed below, the slope deserve to be calculated together follows:slope=fracriserun=fraca_2-a_1t_2-t_1=fracDelta aDelta t
Representation the jerk in graph

### Area Under the Acceleration Graph

The area under the acceleration graph represents the readjust in velocity. In other words, the area under the graph because that a particular time interval is equal to the adjust in velocity during that time interval.Area=Delta VLet us consider the below example to understand better:

The graph below shows a consistent acceleration of 4 m/s2 for a time of 9 s.

Acceleration is identified as,

Delta a=fracDelta vDelta t

By multiply both sides of the equation by readjust in time Delta t, we get

Substituting the values in the above equation, us get

Multiplying the acceleration by the time interval is indistinguishable to recognize the area under the curve.

The area under the curve is a rectangle. This area have the right to be found by multiplying height and also width.

The height, in this case, is 4 m/s2 and the broad is 9 s.

area=4,m/s^2 imes 9,s=36,m/s

The area under any type of acceleration graph because that a details time interval offers the adjust in velocity for the time interval.

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## Acceleration Graph resolved Example

A racecar driver is cruising in ~ a constant velocity that 20 m/s. As she nears the finish line, the race automobile driver starts come accelerate. The graph shown listed below gives the acceleration the the race auto as it starts to rate up. Assume the race car had a velocity that 20 m/s at time t=0 s. Find the last velocity of the driver when she get the end up line.

Solution:

We can discover the readjust in velocity by detect the area under the acceleration graph.

Delta v = area=frac12)(8,s)(6,m/s^2)=24,m/s

Substituting the values, we get

Delta v = area=frac12)(8,s)(6,m/s^2)=24,m/s

This calculation provided us the readjust in velocity throughout the provided time interval. To calculate the last velocity, we need to use the an interpretation of adjust in velocity.

Delta v=v_f-v_i

Substituting the worths in the equation, we get

v_f-20,m/s=24,m/sv_f=44,m/s

Therefore, the final velocity that the racer is 44 m/s.

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Watch the video clip and find out the velocity-time graph and displacement-time graph in detail. Through the end of this video, friend will learn what do the steep of the displacement-time graph and velocity-time graph signify.