Question

under what condition displacement is equal to distance? give a reasonable answer...plz

Answer 1

for a particle at point (x,y) from the origin in usual x-y space, its distance from the origin is \(\sqrt{x^2 + y^2}\) whereas its displacement is \( x ~ \hat i + y ~ \hat j\)

they're fundamentally very different ideas. distance is a scalar, displacement a vector.

even if you make it into a particle moving only along the x axis, you get problems

the displacement is \(\int \dot x (t) ~ dt\)

however the "distance travelled" is \(\int | \dot x (t) | ~ dt\). so a particle oscillating back and forward may have travelled a huge distance as it once again passes the origin where displacement = 0.

messy, ... maybe you have a specific problem you are looking at?

Answer 2

I agree that displacement is a vector while distance is a scalar. That's how we talk about them. As such, they can't be compared directly.

What you might want to know is under what condition the magnitude of the displacement is equal to the distance for a path.

Then you should understand the basics. Walk in an S-shaped path. Distance traveled is the length of the path. Maybe it was 30 meters. Displacement makes you look at two points with one starting point and one ending point. Then you can draw an arrow from the starting point to the ending point and understand both \(\it direction\) and \(\it magnitude\), where magnitude is sort of like the distance between points.
While the distance traveled might be 30 meters, the displacement could vaguely be described as "7 meters in the direction that's up and to the right."

The magnitude of displacement is the distance between points, which you would measure in a straight line because that's how you measure things. So! If the path is a straight line, that path will have the same displacement magnitude as distance. Really, not moving at all gives the same magnitude of displacement as distance, too.