الشارح للدرس: Velocity–Time Graphs | نجوى الشارح للدرس: Velocity–Time Graphs | نجوى

الشارح للدرس: Velocity–Time Graphs الرياضيات

In this explainer, we will learn how to calculate the displacement or acceleration of a particle moving in a straight line from its velocity–time graph.

Imagine an object that moves at a constant velocity, 𝑣, for a period of time that lasts from 𝑡=0s to 𝑡=10s.

A graph of the velocity of the object against time might look like as follows.

Since the velocity is constant, the graph shows a straight, horizontal line. What is the change in the displacement of the object over the time interval shown? Recall that for an object that moves at a constant velocity, 𝑣, for a time interval Δ𝑡, the change in the displacement of the object, Δ𝑠, is given by the formula Δ𝑠=𝑣Δ𝑡.

The object in the graph above moves at a velocity of 12 m/s for a time of 10 s so it moves a distance of 12/×10=120mssm in the direction of its motion. Notice how this is also equal to the area between the line and the 𝑡-axis.

Another question we could ask is “What is the acceleration of the object in the time interval shown?” Since the object is moving at a constant velocity, we know that its acceleration is zero, so we can say that when the line on a velocity–time graph is horizontal, the acceleration is zero.

Now imagine an object that moves at a steadily increasing velocity for a period of time that lasts from 𝑡=0s to 𝑡=10s. A graph of the velocity of the object against time might look like as follows.

Since the velocity is increasing steadily over time, the graph shows a straight, but not horizontal, line. Since the velocity is not constant, we cannot use exactly the same formula that we used before to find the change in the object’s displacement. However, we could use the same formula if we were to find the average velocity of the object, 𝑣average. In this case, the average velocity of the object is just equal to the initial velocity, 𝑣i, plus the final velocity, 𝑣f, divided by two: 𝑣=𝑣+𝑣2.averageif

In this case, the initial velocity is 12 m/s and the final velocity is 18 m/s, so the average velocity is equal to (12/+18/)2=15/msmsms. We can now use the formula for the change in the displacement of the object that we used before: Δ𝑠=𝑣Δ𝑡Δ𝑠=15/×10Δ𝑠=150.averagemssm

But another way of thinking about this is in terms of the area between the line and the 𝑡-axis, or the area under the line. The change in the displacement is equal to the area under the line. We can find the area under the line more easily by dividing it into two regions, as follows.

The blue region is a rectangle, so its area is equal to its height, 12 m/s, multiplied by its width, 10 s, which is 120 m. The red region is a right triangle, so its area is equal to its height, 6 m/s, multiplied by its width, 10 s, divided by two, which is 30 m. So the total area is 120+30=150mmm.

Since we said that the velocity is increasing steadily over time, the acceleration of the object is constant, so we can say that when the line on a velocity–time graph is straight, the acceleration is constant. Since the line slopes up to the right, the object is accelerating, but if it slopes down to the right, it would be decelerating.

Let’s have a look at some examples.

Example 1: Finding the Change in Displacement of an Object over a Given Time Interval from a Velocity–Time Graph

The given velocity–time graph represents a particle moving in a straight line. Determine its displacement at 𝑡=2s.

Answer

This graph shows us the velocity of the particle over 7 seconds, but we are only interested in the first two seconds. The line on the graph is straight and slopes up to the right; this means that the object has a constant, nonzero acceleration.

To find the change in the displacement of the particle between 0 s and 2 s, we just need to find the area between the line and the 𝑡-axis between 𝑡=0s and 𝑡=2s.

Since this region is a triangle, the area is equal to the change in velocity, Δ𝑣, times the time interval, Δ𝑡, divided by two: areaareaarea=Δ𝑣Δ𝑡2=30×22=30.

This area is equal to the change in the displacement of the particle. Since the unit of the velocity was centimetres per second and the unit of time was seconds, this value for the displacement is in centimetre, so the change in the displacement of the particle is 30 cm.

So far, we have only considered positive velocity. If the velocity of the object were negative, however, and the magnitude of the velocity increased steadily with time, a graph of its velocity would look as follows.

In this case, since the velocity is negative throughout the time interval shown, the change in the displacement throughout that time interval is also negative. We can again find the change in displacement by finding the area of the shaded region on the graph, but we then have to multiply that area by 1 to get the change in the displacement: areaarea=10×20×12=100, but since the velocity is negative, we have to multiply this by 1: Δ𝑠=100.m

So we have now considered what we have to do if the velocity is negative throughout a time interval, but what about if the velocity changes between positive and negative? The graph below shows the velocity of an object that starts out positive before becoming negative.

The line crosses the 𝑡-axis at 𝑡=10s; the velocity goes from positive to negative; this is when the object reverses direction. The object starts out moving in one direction and then starts moving in the opposite direction.

In the first ten seconds of its motion, the displacement of the object is increasing, but in the last ten seconds, since the object is moving in the opposite direction, its displacement must be decreasing. This means that in order to find the change in the displacement over the whole time interval from 𝑡=0s to 𝑡=20s, we have to subtract the area between the line and the 𝑡-axis that is below the 𝑡-axis (the red region) from the area between the line and the 𝑡-axis that is above the 𝑡-axis (the blue region). In this case, since the area of the red region is equal to the area of the blue region, the change in the displacement of the object over this 20-second time interval would be zero.

Let’s have a look at another example.

Example 2: Finding the Displacement of a Particle Using a Velocity–Time Graph

Given the velocity–time graph of a particle moving in a straight line, determine the displacement of the particle within the time interval [0,9].

Answer

The question asks us about the displacement between 𝑡=0s and 𝑡=9s, so we can ignore everything on the graph after 𝑡=9s.

Between 𝑡=0s and 𝑡=1s, the velocity is positive, so the change in the displacement in that time interval is also positive. Between 𝑡=1s and 𝑡=9s, the velocity is negative, so the change in the displacement in that time interval is also negative. In order to find the displacement of the particle, we need to subtract the area between the line and the 𝑡-axis for the interval [1,9] from the area between the line and the 𝑡-axis for the interval [0,1].

Let’s start by finding the area between the line and the 𝑡-axis for the interval [0,1]. This region is a triangle, so its area, 𝐴, is given by 𝐴=4×1×12𝐴=2.

Now let’s find the area between the line and the 𝑡-axis for the interval [1,9]. This region is a trapezoid, so its area, 𝐴, is given by 𝐴=8+52×4𝐴=26.

Then, the displacement is given by Δ𝑠=𝐴𝐴Δ𝑠=226Δ𝑠=24.

Since 𝑣 was given in metres per second and 𝑡 was given in seconds, this value for the displacement is in metres.

So far, we have only looked at how we find the displacement from a velocity–time graph. To find the displacement, we add up the areas between the line and the 𝑡-axis that are above the 𝑡-axis and subtract from that all the areas between the line and the 𝑡-axis from below the 𝑡-axis.

However, if instead we wanted to find the total distance traveled by an object from its velocity–time graph, we would instead add all of the areas between the line and the 𝑡-axis, whether or not they are above or below the axis.

Example 3: Finding the Distance Covered by a Particle Using a Velocity–Time Graph

Given the velocity–time graph of a particle moving in a straight line, determine the distance covered by the particle within the time interval [0,8].

Answer

This question only asks us about the time interval [0,8], so we can ignore everything on the graph after 𝑡=8s.

To find the total distance traveled, we just need to add the areas between the line and the 𝑡-axis between 𝑡=0s and 𝑡=8s.

The region between the line and the 𝑡-axis in the interval [0,1] is a triangle, so its area, which we will call 𝐴, is equal to its height multiplied by its width divided by two: 𝐴=5×12𝐴=2.5.

The region between the line and the 𝑡-axis in the interval [1,8] is a trapezoid, so its area, which we will call 𝐴, is equal to its mean width multiplied by its height: 𝐴=4+72×5𝐴=27.5.

The total distance traveled is equal to the total area: totalareatotalareatotalarea=𝐴+𝐴=2.5+27.5=30.

So the total distance traveled by the object in the time interval [0,8] is 30 m.

When the line on a velocity–time graph is straight, we can also work out the acceleration of the object. Over the time interval for which the line is straight, the acceleration is constant and is equal to the change in velocity, Δ𝑣, divided by the time interval, Δ𝑡: 𝑎=Δ𝑣Δ𝑡, where 𝑎 is the acceleration.

Example 4: Finding the Acceleration of an Object from Its Velocity–Time Graph

Given the velocity–time graph for a particle that moved in a straight line, determine its acceleration at 𝑡=3s.

Answer

The line is straight over the entire time interval shown, so the acceleration is constant throughout the time interval.

We can use the formula 𝑎=Δ𝑣Δ𝑡, where 𝑎 is the acceleration, Δ𝑣 is the change in the velocity over the time interval, and Δ𝑡 is the time interval, to find the acceleration of the object.

From the graph, we can see that the velocity of the object increases from 0 cm/s to 210 cm/s, and the time interval is 10 seconds long, so 𝑎=21010𝑎=21, and since the acceleration is constant throughout the time interval shown on the graph, the acceleration of the object at 𝑡=3s is 21 cm/s2.

Example 5: Finding the Acceleration of an Object from Its Velocity–Time Graph

The figure shown is a velocity–time graph for a body moving in a straight line. Determine the deceleration of the body during the final section of its movement, given that it came to rest 100 seconds after it started moving.

Answer

We are only interested in the final section of the object’s motion, which is between 90 s and 100 s, and we can ignore everything outside of that.

In the time interval [90,100], the line on the graph is straight, so the acceleration is constant. That means that we can use the formula 𝑎=Δ𝑣Δ𝑡, where 𝑎 is the acceleration, Δ𝑣 is the change in the velocity over the time interval, and Δ𝑡 is the time interval, to find the acceleration of the object.

From the graph, we can see that the velocity of the object decreases from 45 m/s to 0 m/s, which is a change of 45 m/s, and the time interval is 10 seconds long, so 𝑎=04510090𝑎=4510𝑎=4.5.

The acceleration of the object in the last part of its motion is 4.5 m/s, which is the same as a deceleration of 4.5 m/s.

Key Points

  • The change in the displacement of an object can be calculated from a velocity–time graph by adding the areas between the line and the 𝑡-axis that are above the 𝑡-axis and subtracting the areas that are below the 𝑡-axis.
  • The distance traveled by an object can be calculated from its velocity–time graph by adding all the areas between the line and the 𝑡-axis.
  • When the line on a velocity–time graph is straight, the acceleration can be found using the formula 𝑎=Δ𝑣Δ𝑡.

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