The given velocity-time graph represents a particle moving in a straight line. Determine its displacement at 𝑡 equals two seconds.
There’re two ways of approaching this problem. Firstly, we could find the area under the graph. This is because in a velocity-time graph, the distance is equal to the area under the graph. The second method would be using the equations of motion or suvat equations. We will look at this later. We were asked to calculate the displacement at 𝑡 equals two seconds. Therefore, we need to work out the area between 𝑡 equals zero and 𝑡 equals two.
The area of any triangle can be calculated by multiplying the base by the height and dividing by two. In our triangle, the base is equal to two. And the height is equal to 30. Two multiplied by 30 is equal to 60. So the area is equal to 60 divided by two. This is equal to 30. Our units for velocity were centimetres per second. And the units for the time were seconds. This means that our units for distance or displacement will be centimetres.
The displacement at 𝑡 equals two seconds is 30 centimetres.
Our second method, as mentioned earlier, is to use the equations of motion or suvat equations. The letters 𝑠, 𝑢, 𝑣, 𝑎, and 𝑡 stand for the displacement, initial velocity, final velocity, acceleration, and time, respectively. In this case, we’re looking at the period of time between 𝑡 equals zero and 𝑡 equals two. This means that our time is equal to two seconds. The initial velocity when 𝑡 equals zero was zero centimetres per second. The final velocity when 𝑡 equals two was 30 centimetres per second.
We were asked to calculate the displacement 𝑠. The equation that includes 𝑠, 𝑢, 𝑣, and 𝑡 is 𝑠 equals 𝑢 plus 𝑣 divided by two multiplied by 𝑡. Substituting in our values gives us zero plus 30 divided by two multiplied by two. Zero plus 30 is equal to 30. So we have 30 divided by two multiplied by two. 30 divided by two is equal to 15. Multiplying this by two gives us 30.
Once again, we have proved that the displacement is 30 centimetres.