# Distance != Displacement

The distance covered by jockey between 1st and 6th second if the position is given by !!s=t^2-6t+12!!

- 5m
- 13m
- 7m
- None

**Solution**

This question is a simple one but many answered it incorrectly.

Even when moving along a straight line distance need not be same as displacement.

Displacement is not equal to distance if the jockey changes his direction of movement during the time observed.

The velocity of the jockey is !!v=2t-6m//s!!. So at !!t=1s!! !!v=-4m//s!! and at !!t=6s!! !!v=6m//s!!. This means that the velocity is reversed at a time between !!1s,6s!!. In fact velocity changes its direction of motion at !!t=3s!!.

There are 2 methods to solve it:

**First:** We find the distance covered in each second add it over the next 5 seconds. This works because in a second there is no change in direction for the jockey.

We get that distance travelled in !!2nd!!, !!3rd!!, !!4th!!, !!5th!! and !!6th!! seconds are !!3m,1m,1m,3m,5m!!. So the total distance !!=3+1+1+3+5=13m!!

**Second:** A shorter approach would be to find the distance covered between before turning and after turning.

!!s(1)=7m!!

!!s(3)=3m!!

!!s(6)=12m!!

So the total distance covered is !!|s(3)-s(1)|+|s(6)-s(3)|=13m!!