Math problem please help!!?

Question

Kathryn and Carl are in a ski race. They begin on the start ramp then ski around each of 17 gates and then head for the finish line. The 17 gates befroe the finish line are each spaced 50 meters apart from start to finish. Kathryn begins skiing her race, downhill at 10 kilometers per hour. Carl starts 30 seconds later, skiing at 20 kilometers per hour. Can he catch Kathryn before she crosses the finish line? If so, where will he pass her?

Answer 1

If acceleration (speeding up, slowing down, and turning) is factored in, you do not have enough information to solve the problem. Therefore, I will disregard acceleration.

Neglecting acceleration...(scroll ALL THE WAY DOWN for JUST THE ANSWER)

Assuming the first gate is 50 meters away from the start, and the last 50 meters from the finish, there will be 18 segments (17 gates + 1 start +1 finish = 19 bounds = 18 segments) times 50 meters equals a 900 meter race

Kathryn goes 10,000 meters per hour. To see how long it will take her, set up a proportion. x/900=3600 (3600 secs/hr)/10000

10000X=3,240,000
X=324

Woohoo, Kathryn takes 324 seconds to finish the race.

Now here's where we're lucky. If he goes twice as fast as Kathryn, he'd take one half of the time, or 162 seconds. Adding the 30 second handicap, he finishes in 192 seconds.

So he'll pass her.

Find how many meteres per second each travels...

K=2.778
C=5.556 (remember, when calculating use 162 seconds-since that's how long he can go down in)

In the thirty seconds lost, Kathryn will have traveled 83.34 meters.

Now, think of it this way-in every second, Carl is MATCHING Kathryn's 2.778 meters, and making up 2.778 of the difference (83.34).
83.34/2.778=30. it will take him thirty seconds of HIS RACING TIME to catch up to her. Kathryn has been racing for 60 seconds. If we did it right, the numbers should agree, and they do (2.778*60 does indeed equal 5,556*30)

So, multiply through either one of them and you'll get the answer. WOO HOO! Man, this has taken me forever (I started when your queston was ten minutes old). [EDIT-yeop half of an hour!]

THE ANSWER:::

Carl will pass Kathyrn, 166.68 meteres into the race.

Answer 2

10 km/h = 10,000m/(3600s) = 2 7/9 m/s. In 30 s Kathryn gets a 83 1/3 m lead. Carl, skiing at twice Kathryn's speed, eats up her lead at the rate of 2 7/9 m/s, so it will take him (83 1/3 m)/(2 7/9 m/s) = 30 s to catch her. In that time he'll go 166 2/3 m, which will put him 1/3 of the way between gate 3 and gate 4. That's the basic logic. In algebraic equations, let t be the time it takes Carl to catch up. When he does, they'll both have skied the same distance, so

[rate K][time K] = [rate C][time C]
(25/9)(t + 30) = (50/9)t
(25/9)t + 83 1/3 = (50/9)t
83 1/3 = (25/9)t
30 = t.

Answer 3

17 gates each followed by 50 m plus 50 at the beginning makes a 900 m race. 10 km per hour is 10,000 m per hour, so it would take Kathryn .09 h to do the course. 60 min x .09 = 5.4 minutes. At 20 km it would take Carl half that or 2.7 minutes but add .5 minute for the 30 seconds delay, so 3.2 minutes. He will definitely pass her. Let time = t in seconds when Kathryn takes off. Now 10 km per hour is 2.78 meters per second and 20 kmph is 5.56 mps. 2.78 t = 5.56 ( t - 30 ). Solve this and get 60 seconds after Kathryn takes off.

Yes?

Answer 4

In 30 secnds kathryn goes (10X30)/(60X60) or 1/12 km
Their relative speed=20-10=10 kmPH
Therefore Carl will catch up with kathryn in (1/12)/10 or 1/120 hours or 30 seconds.

Answer 5

Since he is going twice her speed, and she has a 30 second head start, he will overtake her 60 seconds after she started the race, but 30 seconds after he started the race, correct? If this is indeed true, taking into conisderation that 10 k per hour equals 10000 m per hour and 166-2/3 m per minute, the spot he passes her should be at exactly that spot...166-2/3 meters down the hill.

Does that make it easier to understand?

Answer 6

It will take Carl 1 min. to catch Kathryn. He will catch her between the 3rd and 4th gate at the 167m. mark.

Answer 7

we should consider force due to gravity, the friction between ski ramp and the ski, the angle of inclination of the ski ramp.
it is more of a physics question.

Answer 8

ouch that is hard. doesn't sound like a math problem, more like physics to me.