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 2.0 m/s, skates by with the puck. After 2.80 s, the first player makes up his mind to chase his opponent. If he accelerates uniformly at 0.32 m/s^2, 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.) 5.91 X Express the position of each player as a function of time. Use that expression to find the time when the two players meet. Make sure you are correctly taking into account just when the second hockey player begins accelerating. s (b) How far has he traveled in that time? m

Explanation / Answer

a)

here

speed = distance / time

distance x = 2 * 2.8 = 5.6 m

then by using the second equation of motion

x = 5.6 + v * t = 0.5 * a * t^2 = 0.16 * t^2 = 5.6 + 2 *t

0.16 t^2 - 2 t - 5.6 = 0

by solving this quadratic equation we get

t= 14.85 , -2.35 sec

so we take the positive value of t = 14.85 sec

b)

by using the second equation of motion

y = ut + 0.5 * a* t^2

y = 0.5 * 0.32 * 14.85^2 = 35.28 m

the distance traveled is 35.28 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.