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