Physics - Solved, thank you.

[godspeedx]

How do you do this problem? I don't necessarily need the answer, I'm just not sure how I can do it. The only formula we know is V = X/T. (velocity = distance/time)

"Two cars are traveling in the same direction along a straight highway, one going 55mi/h constantly the other going 70 mi/h constantly. How far must the faster car travel before it has a 15 min lead on the slower car?"

to save some time:
Car A is traveling at 24.6 meters/second
Car B is traveling at 31.3 meters/second
15 minutes = 900 seconds

Thanks.

 

[Anubis]

just put 15min in for t and solve. its easy

 

[Wheatmaster]

this aint the homework hotline.
times 900 to both 24.6 and 31.3, get the difference.

 

[Hanpan]

THink of this. How far ahead must the faster car be for the distance to be equal to that travelled in 15 mins by the slower care.

Also remeber that you can think of the first car as standing still with respect to the second car. So ask yourself if this is the case how fast is the first faster car moving with respect to the now stationary car.

 

[godspeedx]

Ooooooooh, I was thinking of it in a muuuch harder way. Thanks!

Wait. . . there's something wrong.

 

[Krakerjak]

assume the slower car is stationary, and the faster car is traveling at the difference in speed between the 2 cars
ie. 15m/h

use v=x/t

you have v and t already.....solve for x

 

[godspeedx]

Originally posted by: Krakerjak
assume the slower car is stationary, and the faster car is traveling at the difference in speed between the 2 cars
ie. 15m/h

use v=x/t

you have v and t already.....solve for x

Ok, now it seriously clicked. I wasn't using the difference in the speed. That's what was confusing me.

 

[silverpig]

Originally posted by: Krakerjak
assume the slower car is stationary, and the faster car is traveling at the difference in speed between the 2 cars
ie. 15m/h

use v=x/t

you have v and t already.....solve for x

Unfortunately that's wrong dude.

 

[godspeedx]

Originally posted by: silverpig

Originally posted by: Krakerjak
assume the slower car is stationary, and the faster car is traveling at the difference in speed between the 2 cars
ie. 15m/h

use v=x/t

you have v and t already.....solve for x

Unfortunately that's wrong dude.

Oh yeah that is wrong, because then you would only get the distance of the car traveling at 6.7 meters/second (the difference in the speeds). He's really traveling at 31.3 m/s.

Darn lol, then what is the right way?

 

[godspeedx]

bump

 

[Krakerjak]

i give up

 

[silverpig]

What you do is you want the faster car to have a 15 minute lead on the slower car. This does NOT mean that you just make the cars travel for 15 minutes. I'll give you an example to see why.

Let's say car A is the concorde jet, and car B is you walking. Now let them go for 15 minutes, then stop them (as is suggested by some in this thread). Now, tell me how you can say that the concorde is 15 minutes ahead of you. It's like, WEEKS ahead of you because it's so much faster. The 15 minutes ahead doesn't refer to the concorde's speed, or the difference in speed, but it refers to how much time it will take you to catch up to it. This is why all suggestions so far in this thread are wrong.

A fifteen minute lead on the slower car is a distance that the slower car travels in 15 minutes. Let this distance be called Xa

Xa = Va*T1

Where Va and T1 are the velocity of the slower car, and the time it travels (900 seconds).

Xa = 24.6*900 = 22140 m

So this is the lead that the faster car must generate over the slower car. NOW you can use the other suggestions in the thread and take the difference in velocities. Let Vd = Vdifference between the cars = Vb - Va. Taking T2 = Xa/Vd you will get the time it takes for the faster car to generate this difference.

Vd = 31.3 - 24.6 = 6.7 m/s

and now

T2 = 22140/6.7 = 3304.5 s

So this is the time that it takes the fast car to generate the 15 minute lead. To figure out how far it has travelled, you just plug it back into v = x/t as you have a t and a v.

x = v*t = 3304.5 * 31.3 = 103430.85 m or ~103 km

 

[godspeedx]

You sir, are correct.
Thanks man.