A hockey player is standing on his skates on a frozen pond when an opposing play

Question

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 10 m/s, skates by with the puck. After 4.0 s, the first player makes up his mind to chase his opponent.

(a) If he accelerates uniformly at 3.0 m/s2, how long does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.)

b) How far has he traveled in this time?

So the first skater had already traveled 40m (10m/s in 4s) and because his speed is constant, his acceleration in 0 m/s2.

Their final positions will be equal and the first skater`s time minus 4 seconds will be the second skater`s time.

x final = x initial + v initial (t) + (a(t2))/2

x final = x inital + .5(v inital + v final)t

Skater 2 = x final= 0m + 0t + (3m/s2 (t2))/2
Skater 1 = x final= 40m +.5 (10m/s + 10m/s)t
Skater 1 = x final = 40m + .5(20m/s)t

Since x fial is the same for Skater 1 and Skater 2

40m +.5(20m/s)t = ((3m/s2)(t2))/2

40m + (20m/s)t = 3 m/s2 * t2

0= 3 m/s2 * t2 - 20 m/s * t - 40m

Use the quadratic and get 64.5 and -24.5 for t

So then t would be 64.5 for Skater 1 and (64.5 -4) 60.5 for Skater 2.

Skater 2= v final = v inital + at
Skater 2 = v final= 0m/s + 3m/s2 (60.5s
Skater 2 = v final= 181.5 m/s

Skater 2 =x final = x inital + .5(v inital + v final)t
Skater 2 = x final = 0m + .5(0m/s + 181.5 m/s)(60.5 s)
Skater 2 = x final = .5(181.5 m/s)(60.5 s)
Skater 2 = x final = 5490.4 m

My anwers don`t make ANY sense! Plaese help!

Explanation / Answer

Great job setting everything up. One very simple mistake comes two lines after you wrote,
"Since x fial is the same for Skater 1 and Skater 2"
You doubled the second and third terms (with time in them), but you forgot to double the first term, leaving it as 40m.
So, the good news items are that you did all the hard parts correctly and you recognised that there was something wrong. The bad news is that it is really hard to catch simple mistakes.

Once you change the first term to 80m the quadratic eqn gives you
t = (20 +- 36.878)/6
The negative number tells you they could have met before the problem started (makes no sense in this problem). The positive time, 9.479 s is how long, after the four second wait, they will meet. Since they meet when they have both traveled the same distance, lets use the constant velocity skater. He has gone 40m +94.79m or 135m.
So the answers are (one significant figure?) a) 13 seconds after the problem starts or 9 seconds after he starts the chase. and b) 135 m.

Always check your work, does
40m + 10m/s 9.479s = 1.5 m/s2 (9.479s)2 ?

Yes 134.7 = 134.7.

Does it seem right? Yes about ten seconds accelerating at 3 m/s every second means you are going 30m/s when you catch the other skater, so your average speed was 15 m/s. Average speed times time is 15 m/s times 10 s = 150 m. Meanwhile the other guy went 40m + 10s * 10 m/s = 140 m, so estimate shows that answer looks good.

One note: the skater who is accelerating is going really fast at the end. Can skaters really skate that fast? Look it up and see. The world record for 1,500m is an average speed of 15 m/s

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