Simple Relative Motion Police Car Problem

[mwells]

Ok, here’s the question:

A motorist traveling at 89 km/hr is being chased by a police car at 115 km/hr. If the police car starts from 3.5 km back, how long does it take to catch the motorist? Leave time in hours.

Now I’m looking at this as a relative motion problem since there is no acceleration, therefore time should be equal to the distance between the cars, divided by the difference in the velocity between the two cars.

Therefor; t = (3.5 km) / [(115 km/hr) - (89 km/hr)]

Therefore; t = 0.13 hr

However it’s apparently wrong, so does anybody know what I’m doing wrong and how to fix it? Thanks

 

[recon]

I don't see what is wrong with this.

 

[Enos]

M moves 1.5km every minute
P moves 1.9km every minute
M moves 0.02km per second
P moves 0.03km per second

M+8mins = 15.4
P+8mins = 15.4

M x 8 = 11.9 + 3.5 = 15.4
P x 8 = 15.3
-------------------------------------

M+8mins = 15.4 + m4secs = 0.1 = 15.5
P+8mins = 15.3+ p4secs = 0.12 = 15.5

I did this the long way because this is the first time I tried a question like this.
so I get 8mins and 4 secs with a total distance of 15.46kms
Does it sound right?

But just because it takes that much time to be at the same distance, the question is whether or not the motorist stops.


EDIT I truncated your spurious digits! You must learn to use only the digits which have meaning!
Integral

 

[Sirus]

Yes, that is the same answer as mwells got using his method. Mwells, I think there's a mistake in your solutions manual or where ever the answer is from. Your answer is correct.

 

[Enos]

What method is that? I would love to understand it.

 

[Nenad]

Ill give you a hint, and I am drunk right now, so that makes this problem easy. Use relative velocity. Subtrqact their welocities from each other and then use the equation $$d = {V_o}t + \frac{1}{2} at^2$$

 

[tony873004]

Ill give you a hint, and I am drunk right now, so that makes this problem easy. Use relative velocity. Subtrqact their welocities from each other and then use the equation $$d = {V_o}t + \frac{1}{2} at^2$$

Except he needs t to solve this formula. But t is what he's looking for. a = 0 in this formula, so everything after the + disappears, and he's left with the simple d=vt formula which can be rewritten as t=d/v. But he needs d to solve this.

I also get the same answer. Is this one of those computer-graded homework assignments? It might not like your significant digits. It may or may not want you to put the word "hours" after your answer.

 

[Integral]

Let d be the distance traveled by the police car.
Let V_p = the speed of the police car
Let D be the distance traveled by the fleeing car.
Let V_c = the speed of the fleeing car.

We must have D + 3.5 = d and the time of travel (t) for each car be the same.

D = V_ct
d=V_pt

V_ct+3.5 = V_pt

Now solve for t,

$$t = \frac {3.5} {V_p - V_c}$$

This is the solution given in the first post. As suggested you may need to play with the number of significant digits. (should be 2)