Word Problem help: ...bank robber has had a 3 minute head start...

[billythekid]

Hi all. Hope your Sunday is going well. I am having some issues forming an equation from a word problem. The word problem is over uniform-motion.
The base formla is d=rt.

An police officer is traveling at a rate of 90mph to reach a bank robber. The bank robber has had a 3 minute head start. It will take the officer 12 minutes to overtake h robber. I am to find the speed at which the robber is traveling.

The issue I am having is forming this into the actual equation. Not solving it.

I have tried:

r=90(r*3/10)
/
12

r=12(r*3/10)
90

r=90
/
12(r*3/10)

All of which fail to solve to the correct answer.

Thanks in advance.

 

[Johulus]

Hi all. Hope your Sunday is going well. I am having some issues forming an equation from a word problem. The word problem is over uniform-motion.
The base formla is d=rt.

An police officer is traveling at a rate of 90mph to reach a bank robber. The bank robber has had a 3 minute head start. It will take the officer 12 minutes to overtake h robber. I am to find the speed at which the robber is traveling.

The issue I am having is forming this into the actual equation. Not solving it.

I have tried:

r=90(r*3/10)
/
12

r=12(r*3/10)
90

r=90
/
12(r*3/10)

All of which fail to solve to the correct answer.

Thanks in advance.

So...
\(\displaystyle v_{p} .... the \quad speed \quad of \quad the \quad police \\ v_{r} .... the \quad speed \quad of \quad the \quad robber \\ t_{p} .... time \quad required \quad for \quad the \quad police \quad to \quad get \quad to \quad robber \\ t_{r} ..... total \quad time \quad that \quad robber \quad has \quad been \quad driving \)
\(\displaystyle v_{p}=90 mph \\ t_{p}= 12 minutes =\dfrac{1}{5} h \quad ( have \quad to \quad be \quad same \quad units \quad as \quad in \quad speed) \\ t_{r}= t_{p} + 3 minutes= \dfrac{1}{4} h\\ v_{r}=? \)
\(\displaystyle S_{p}=S_{r} \quad (The \quad total \quad distance \quad that \quad both, \quad the \quad robber \quad and \quad the \quad police, \quad cross \quad is \quad equal....
\\ v_{p}\cdot t_{p} = v_{r} \cdot t_{r} \)

 

[stapel]

An police officer is traveling at a rate of 90mph to reach a bank robber. The bank robber has had a 3 minute head start. It will take the officer 12 minutes to overtake h robber. I am to find the speed at which the robber is traveling.

The issue I am having is forming this into the actual equation. Not solving it.

I have tried:

r=90(r*3/10)
/
12

r=12(r*3/10)
90

r=90
/
12(r*3/10)

All of which fail to solve to the correct answer.

What was the reasoning by which you arrived at these equations?

The cop and the robber start at the same place (the robbery site), and end at the same place (the capture site), so their distances are the same.

How long does the cop take, in hours? What is his speed? So how far did he travel?

How long did the robber take, in hours? (Hint: Subtract.) What was his distance? Then what was his rate?

If you get stuck, please reply showing your work and answers to the above questions. Thank you!

 

[billythekid]

So...
\(\displaystyle v_{p} .... the \quad speed \quad of \quad the \quad police \\ v_{r} .... the \quad speed \quad of \quad the \quad robber \\ t_{p} .... time \quad required \quad for \quad the \quad police \quad to \quad get \quad to \quad robber \\ t_{r} ..... total \quad time \quad that \quad robber \quad has \quad been \quad driving \)
\(\displaystyle v_{p}=90 mph \\ t_{p}= 12 minutes =\dfrac{1}{5} h \quad ( have \quad to \quad be \quad same \quad units \quad as \quad in \quad speed) \\ t_{r}= t_{p} + 3 minutes= \dfrac{1}{4} h\\ v_{r}=? \)
\(\displaystyle S_{p}=S_{r} \quad (The \quad total \quad distance \quad that \quad both, \quad the \quad robber \quad and \quad the \quad police, \quad cross \quad is \quad equal....
\\ v_{p}\cdot t_{p} = v_{r} \cdot t_{r} \)

Thank you Johulus! That set me in the right direction.