complicated exercise in kinematics physics .

Question

A car , C, travelling at constant speed u, passes a stationary police car , P , which immediately sets off in pursuit with acceleration a , find in terms of u and a , the greatest distance separating the two cars during the subsequent chase .

Answer 1

The solution above is wrong, because the derivative was calculated improperly.  The derivative of 1/2at^2 is at.

Note that you don't really need to do this.  The greatest separation will occur when the police car's velocity equals that of the car.  After this, the police car is faster, so the distance is decreasing.  Before, the car is faster than the police car, so the distance is increasing.

With constant acceleration, v of the police car = at.

Thus, at = u, or t = u/a

Then, the maximum distance is ut - 1/2 at^2 = u (u/a) - 1/2 a (u/a)^2 = u^2/a - 1/2 u^2/a = 1/2 u^2/a or u^2/2a

Answer 2

Hi Nona!

 

First, let's get an expression for the magnitude of displacement between the two cars (since they are both presumably going in s straight line, the magnitude of displacement and distance are identical).  The two cars start from the same position at the same time.

 

Car C is traveling at a constant velocity u (in the positive direction, presumably).  So its displacement from the position where it passed the police car is as a function of time, given by:

 

d = vt

 

So, in this case, 

 

d_C = ut

 

The police car is accelerating, and starts from rest, so its displacement is given by:

 

d_P = (1/2)at²

 

Therefore, the distance between the two cars at any point (presumably until the police car catches Car C) is given by:

 

Δd = d_C - d_P = ut - (1/2)at²

 

To find the point in time where this value Δd is at an extreme value, we appeal to calculus (if calculus is not allowed in your solution, then let me know, and we can talk about other ways to solve it):

 

First, take the derivative of Δd with respect to time:

 

(d/dt)(Δd) = u - at

 

Then set that derivative equal to zero and solve for t:

 

u - at = 0 --> t = u/a

 

To demonstrate that this represents a maximum value for Δd, and not a minimum, we take the second derivative of Δd with respect to time, or the derivative of the first derivative.

 

(d²/dt²)(Δd) = -a

 

Since the second derivative of Δd is negative, the extreme value of Δd that we are dealing with above must be a maximum.  So, now, we just plug that value for t back into the Δd expression to solve for the value of Δd at that time:

 

Δd_max = u(u/a)-(1/2)a(u/a)² = u²/a-(1/2)(u²/a) = (1/2)(u²/a) = u²/2a

 

I hope this helps!  As I mentioned, if calculus is not allowed in your solutions, then Tom's solution works very well (and is computationally shorter, in any case).  Also, if you have any questions, just let me know.