Kinematics

Question

A particle is projected upwards from the top of a tower. Treat point of projection as origin & upwards
as positive. Consider three point on its path such that at (s represents displacement from origin and v
represents velocity)

(i) s > 0, v > 0 (ii) s > 0, v < 0 (iii) s < 0, v < 0

Which of the following correctly indicates the distance travelled by particle since it's projection to respective points in appropriate order

(A) (i) > (ii) > (iii)

(B) (i) = (ii) < (iii)

(C) (i) < (ii) < (iii)

(D) ordering depends on exact location of points

Answer 1

Hi, Sam -

 

In looking at the problem, it appears that D is the answer.

 

The case (i) represents those displacements in which the particle is travelling upward because of the positive velocity.

 

Case (ii) represents those displacements in which the particle is travelling downward because the velocity is less than zero (negative).

 

Because there are instances when the displacement in i would be greater than ii (e.g. near the peak of its trajectory going up vs. just before it lands going down) and instances where the displacement in ii is greater than in i (e.g. just as it starts to fall vs. right at the beginning of its climb) you would need to know the position of the particle to correctly order the events.

 

Also, for Case iii, it shows s (displacement) less than 0. I am not sure if that is possible since the particle should always be some distance away from the origin or at the origin (which is 0).