Distance and Displacement from a velocity vs. time graph

My teacher is saying that the distance covered will be equal to the area of the trapezium in the graph, but the displacement will be equal to the area of the triangle (with purple hypotenuse). I studied in the internet, and it seemed to me that in this case, the distance will be equal to the displacement (as the velocity was not negative at any point of time). But my teacher is saying otherwise. He is not explaining the reason either. Can anyone please explain it to me? I really need to understand it.

*unit of time is second and unit of velocity is metre per second

• How do you define distance and displacement? – Steeven Feb 10 '18 at 11:15
• There are four different colored lines. What is each one supposed to represent? – Chris Feb 10 '18 at 11:23
• Nothing. I just wanted to provide a clear picture. So I recreated the graph in my question paper in my phone. – Hoque Feb 10 '18 at 11:25

Position is the time-integral to velocity:

$$s=\int v\; dt$$

And the velocity is your red-and-then-green curve according to your text. The purple line is describing another motion.

The total length/distance/displacement (depending of how you define each word) covered over some duration is simply the definite integral (with $t_1<t_2$):

$$s=\int_{t_1}^{t_2} v\; dt$$

And such an integral is the area under the $v$-curve (still the red-and-green curve, not the purple). If you define one of these words as sign-independent, then we must at all times keep a positive speed, and we simply add:

$$s=\int_{t_1}^{t_2} |v|\; dt$$

But, regardless of which version you are actually looking for, we can see that there are no situations of negative $v$-values on this graph. Both formulae are completely equal in this case.

Now, that your teacher should claim the purple line to be the real motion does not make much sense. I am guessing that you must have misunderstood, or the teacher has misspoken.

Imagine two race cars with the same max speed, but one with better acceleration. One will quickly reach it's max speed and thus move further-per-second at every moment than the other one. It will move far ahead. When the other one reaches the same max speed, the first one is already much further ahead. If you stop the race now, then of course it will end up having covered a longer distance, intuitively.

The graphs below show that there can be a difference between the distance travelled and the magnitude of the displacement.

For the motion depicted by the left hand graph the displacement and the distance travelled are both equal to the shaded area $A$.
They are the same because the object is always travelling in the same direction.

For the motion depicted by the centre and right hand graphs the direction of motion of the object changes - its velocity becomes negative after a certain time.

For the centre graph you need to find the difference between the two area $B$ and $C$ to obtain the displacement whereas with the right hand graph you need to add the two areas to obtain the distance travelled.

If areas $B$ and $C$ are the same then the displacement is zero whereas the distance travelled is $B+C = 2B = 2C$ .