Question Video: Determining Displacements at Different Times Using a Velocity-Time Graph | Nagwa Question Video: Determining Displacements at Different Times Using a Velocity-Time Graph | Nagwa

# Question Video: Determining Displacements at Different Times Using a Velocity-Time Graph Physics • First Year of Secondary School

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The velocity–time graph shows the change in the velocity of a person walking in the time interval from 𝑡 = 0 seconds to 𝑡 = 6 seconds. The person’s starting point is their position when 𝑡 = 0 seconds. What is the person’s displacement from their starting point after two seconds?

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### Video Transcript

The velocity–time graph shows the change in the velocity of a person walking in the time interval from 𝑡 equals zero seconds to 𝑡 equals six seconds. The person’s starting point is their position when 𝑡 equals zero seconds. What is the person’s displacement from their starting point after two seconds?

Looking at the graph, we can see velocity on the vertical axis and time on the horizontal axis. The person initially starts walking with a constant velocity, then slows down, comes to zero, and speeds up, becomes negative, so starts walking in the opposite direction, and then carries on with a constant velocity. If we define positive velocity as being towards the right of the screen, then we could imagine this as a person walking towards the right and then slowing down and coming to a stop, before speeding up again and then carrying on at a constant velocity in the opposite direction. If we were to plot this on a displacement–time graph, we would initially have displacement increasing at a constant rate, then turning over, and finally decreasing at a constant rate. So how do we find the displacement from a velocity–time graph?

Recall that displacement is the area under a velocity–time graph. And by “area under” we mean the area between the line and the horizontal axis. Now, remember also that displacement is a vector quantity, meaning it has both magnitude and direction. So this area shaded above the axis indicates positive displacement, and this area shaded under the axis indicates negative displacement.

So let’s now look at the person’s displacement from the starting point after two seconds. For this, we need to calculate the area of this shaded region. And the area of a rectangle is just the horizontal value, which is two, times the vertical value, which is three. And that gives us six. And for the units, we take the units of the vertical axis, which are meters per second, and multiply by the units of the horizontal axis, so that’s time seconds. And then the seconds cancel, and we’re left with a displacement after two seconds of six meters.

Now we need to clear some space on screen, but we’re going to use this value again. So let’s make a small table of time in seconds and displacement after that amount of time in meters. And we can enter that after two seconds, we have a displacement of six meters.

The next part of the question asks, what is the person’s displacement from their starting point after three seconds? Now, we’ve already calculated the area under the line up to two seconds. So we just need to add this additional area between two and three seconds. Now, for a triangle with a base length of 𝑏 and a height of ℎ, the area is equal to half the base times the height. So the displacement from two to three seconds is equal to half times the base of the triangle, which is one, times the height of the triangle, which is three. And a half times one times three is equal to 1.5.

Now we need to add this to the displacement after two seconds, which we’ve already found. So that’s six. And that gives us a total displacement after three seconds of 7.5. And again, that’s in meters. So the person’s displacement from their starting point after three seconds is 7.5 meters. Now let’s add that to the table. At a time of three seconds, we have a displacement of 7.5 meters.

Next, we’re asked what is the person’s displacement from their starting point after four seconds. And again, we’ve already calculated the area up to time of three seconds. So we just need to add in this extra area between three and four seconds. And here we have a triangle the same size as the previous one. But this time it’s below the horizontal axis. So this time we need negative sign. And we have negative one-half times the base of the triangle, which is one, times the height of three. And that gives us negative 1.5.

Now we need to add this to the value we found previously of the displacement up to a time of three seconds, which was 7.5. And we have negative 1.5 plus 7.5, which is equal to six. And again, that’s in meters. So the person’s displacement from their starting point after four seconds is six meters.

For the final part of the question, we need to find what is the person’s displacement from their starting point after five seconds. And here we just need to add in this extra shaded rectangle between four and five seconds. Now, again, this is below the horizontal axis. So we need a negative sign. And we have negative the width of the rectangle, which is one, times the height, which is three, which gives us a displacement of negative three. And to this we’re going to add the value we found previously of the displacement up to four seconds, which was six meters.

And so the total displacement after five seconds is negative three plus six, which is three meters. So the person’s displacement from their starting point after five seconds is three meters.

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