Aproton is projected toward a fixed nucleus of charge +ze with velocity vo. initially the two particles are very far apart. when the proton is a distance r from the nucleus its velocity has decreased to 1/2vo.
how far from the nucleus will the proton be when its velocity has dropped to 1/4vo?
4/5 r
1/4 r
1/2 r
1/16 r

Explanation:

Here, the kinetic and potential energy is conserved

For the second case

Subtract the equations we get

Divide the equations

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.