Kinematics question

Question

Two particles P and Q moves in the positive direction on the x axis, P with constant acceleration 2 ms^-2 and Q with constant acceleration of 1 ms^-2. At time t=0, P is projected from the origin O with speed 1 ms^-1, and at time t=4, Q is projected from O with speed 16 ms^-1.

(a) Find the times between which Q is ahead of P.

(b) Find also the distance from O at which Q overtakes P, and the distance from O at which P overtakes Q.

Answer 1

Since this motion is in one dimension, and the accelerations are both constant, we can use one-dimensional kinematics on this problem.  We are particularly interested in the displacements and times at which the particles are at the same position.  We are given accelerations and initial velocities, and are interested in times and positions.  The one kinematic equation that brings them all together is:

 

x-x_o = v_ot + (1/2)at²

 

where

 

t = time

x = position along axis at time t

x_o = initial position

v_o = initial velocity

a = acceleration

 

Since both particles start from the same position (the origin), their positions (x) will be equal when their displacements (x-x_o) are equal.  We also know Q is launched 4 seconds after P, so whatever the time is in the kinematic equation for P, the equation for Q must have (t-4) at the same time; since, at any instant, Q has been on the move for 4 fewer seconds than P.

 

So we want to determine, first, the time or times at which the particles are at the same position.  With the given information, we can write the kinematic equations for each particle as:

 

(x-x_o)_P = (1 m/s)t + (1/2)(2 m/s²)t²

(x-x_o)_Q = (16 m/s)(t-4) + (1/2)(1 m/s²)(t-4)²

 

To solve for the times when the displacements (and thus the positions) are the same, we can set both right sides above equal to each other, and solve for the value(s) of t that make that equation true.  I will forego putting the units in from now until the end:

 

The right side of the P equation simplifies to: t + t²

The right side of the Q equation simplifies to: 16t - 64 + (1/2)(t²-8t+16) = 16t - 64+(1/2)t² - 4t + 8 = (t²/2)-12t-56

 

Now we set those two right sides equal to each other:

 

t+t² = (t²/2)+12t - 56

 

which, when put into standard quadratic form, becomes

 

(t²/2) - 11t +56 = 0

 

Solving this quadratic equation will give the value(s) of t at which the displacements, and thus the positions, of the two particles are equal.  Note that, by casting the Q equation with times of (t-4), we are now solving for times t referenced to t=0 (the time of P's launch), as the problem requests.

 

Make sure you can do the algebra steps.  I found I was able to factor the above quadratic equation and get values of t = 8 s, and t=14 s.

 

Logically, the earlier time will be when Q (launched second) catches up to P, and the later time will be when P again catches and passes Q.  This can be confirmed by substitution into the kinematic equations (at any time between 8 and 14 s, the displacement of Q should be greater than the displacement of P).  So Q would be ahead of P between 8 and 14 seconds after t=0.

 

To determine the displacements from the origin at which these two passings happen, we just have to solve for the displacement (of either particle; they should be the same) at 8 s and 14 s.  I solved for the displacement of both P and Q at both times, just to confirm the values were the same.  I obtained:

 

displacement at which Q overtakes P (at t = 8 s):  72 m

displacement at which P overtakes Q again (at t = 14 s): 210 m

 

Make sure you can work out the details of how to get all solutions.  If you want to look at it more closely, just let me know.  I am also available for online tutoring help if you have more in-depth questions about this or related subjects.