Speed question

• Feb 7th 2008, 06:54 AM
Raj
Speed question
Runner #1 is 35m ahead of Runner #2.
Runner #1 travels at 1.65m/s and Runner #2 travels at 1.85m/s.
Both run in the same direction.
From where they begin, how far does Runner #2 travel before he catches up with Runner #1, how long does it take?

I know i have to use variables, but i'm lost. Any help please
• Feb 7th 2008, 07:16 AM
colby2152
Quote:

Originally Posted by Raj
Runner #1 is 35m ahead of Runner #2.
Runner #1 travels at 1.65m/s and Runner #2 travels at 1.85m/s.
Both run in the same direction.
From where they begin, how far does Runner #2 travel before he catches up with Runner #1, how long does it take?

I know i have to use variables, but i'm lost. Any help please

You want to find what time (and therefore location) where the runners are at the same spot. Runner #1 is 35 meters ahead of the other one, so use a distance equation...

$\displaystyle d=rt$

$\displaystyle d_1+D=d_2$

$\displaystyle r_1t+D=r_2t$

$\displaystyle 1.65t+35=1.85t$

$\displaystyle 35=0.2t$

$\displaystyle t=175$

At 175 seconds into the "race", Runner #1 and Runner #2 are in the same spot. Plug $\displaystyle t=175$ into the original equation: $\displaystyle 1.65t+35=1.85t$

$\displaystyle 323.75=323.75$
• Feb 7th 2008, 07:17 AM
wingless
You don't have to use any equation or formula for this question. We will just use our imagination :)

(This solution is similar to my post here)

First, imagine the scene as you're on the ground, standing, looking at the runners. You see that they're 35 m away and running at the same direction at 1.85 m/s and 1.65 m/s.

Now, imagine the scene as you are Runner #2. You're 35 m behind Runner #1. You both start running. Ignore the grounds. What do you see? You see that Runner #1 gets closer at 0.2 m/s! (it's 1.85 - 1.65). We just calculated the relative speeds. And now, he's 35 m ahead and coming closer at 0.2 m/s. When would you meet? It's so easy,
$\displaystyle V = \frac{x}{t}$
$\displaystyle 0.15 = \frac{35}{t}$
$\displaystyle t = \frac{35}{0.2}$
$\displaystyle t = \frac{350}{2} = 175$ seconds

Now you work the rest :)
• Feb 7th 2008, 09:19 AM
Raj
Quote:

Originally Posted by wingless
You don't have to use any equation or formula for this question. We will just use our imagination :)

(This solution is similar to my post here)

First, imagine the scene as you're on the ground, standing, looking at the runners. You see that they're 35 m away and running at the same direction at 1.85 m/s and 1.65 m/s.

Now, imagine the scene as you are Runner #2. You're 35 m behind Runner #1. You both start running. Ignore the grounds. What do you see? You see that Runner #1 gets closer at 0.2 m/s! (it's 1.85 - 1.65). We just calculated the relative speeds. And now, he's 35 m ahead and coming closer at 0.2 m/s. When would you meet? It's so easy,
$\displaystyle V = \frac{x}{t}$
$\displaystyle 0.15 = \frac{35}{t}$
$\displaystyle t = \frac{35}{0.2}$
$\displaystyle t = \frac{350}{2} = 175$ seconds

Now you work the rest :)

I really have to remember this way of solving, test in 2 hours so

Thanks (again) (Hi)