Question

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 2.0 m/s, skates by with the puck. After 2.40 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 0.12 m/s2, determine each of the following.

(a) How long does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.)
_____ s

(b) How far has he traveled in that time?
_____ m

Answer 1

a) Use: \(x_f=x_i+v_0\Delta t+\dfrac{1}{2}a(\Delta t)^2\)

guy with puck (B): constant velocity =2.0 m/s, a=0 , \(x_i\)=4.8 m

\(x_f=4.8m+(2.0~m/s)\Delta t\)

guy chasing (A): constant acceleration, initial velocity =0, \(x_i\)=0

\(x_f=\dfrac{1}{2}(0.12~m/s^2)(\Delta t)^2\)

they want to know the time when he catches up, meaning their final positions \(x_f\) are the same. So make them equal to each other:

\(4.8m+(2.0~m/s)\Delta t=\dfrac{1}{2}(0.12~m/s^2)(\Delta t)^2\)

solve using the quadratic formula, you should get \(\Delta t\)=35.6 s \(\approx\) 36 s

b) use the original formula for guy A with the time you have:

\(x_f=\dfrac{1}{2}(0.12~m/s^2)(36~s)^2\)=77.76 m \(\approx\) 78 m

Answer 2

Thanks for the explanation, but after solving the quadratic formula I got Δt= -2.65 s and -30.65 s.

After solving the equation 4.8m+(2.0m/s)t=1/2(0.12 m/s^2)(t)^2 I used t^2+33t+80=0 to solve using the quadratic equation.

I did it a couple times but still got the same answers over again.

Answer 3

i solved it on the computer with this website http://web2.0calc.com/

you shouldve had:

\(4.8+2.0x=\dfrac{1}{2}(0.12)x^2\)

\(0.06x^2 -2x-4.8=0\)

if you divide through with 0.06

\(x^2 -33.333x-80=0\)

so i think you messed up on the signs and rounding down.

Answer 4

ps. you will have 2 solutions, a negative one and a positive one. Use the positive one because negative time doesnt have a real physical value.

Answer 5

Thanks again! I'll try and see if I get it this time :)

Answer 6

sounds like a plan, good luck!