Answer to Question #162514 in Physics for Chrishan

Question

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a 

uniform speed of 14 m/s, skates by with the puck. After 3.5 s, the first player makes up his mind to chase 

his opponent. If he accelerates uniformly at 4.5 m/s2

, (a) how long does it take him to catch his opponent,

and (b) how far has he traveled in that time? (Assume the player with the puck remains in motion at 

constant speed.) 

Answer 1

(a) Let's first find the distance covered by the opposing player in 3.5 second:

$$d=vt=14\ \dfrac{m}{s}\cdot3.5\ s=49\ m.$$

Let's write the position's of both first and opposing players:

$$x_1=x_{01}+v_{01}t+\dfrac{1}{2}a_1t^2,$$ $$x_2=x_{02}+v_{02}t+\dfrac{1}{2}a_2t^2.$$

When the first player catches the opposing player, $x_1=x_2.$ Therefore, we can write:

$$x_{01}+v_{01}t+\dfrac{1}{2}a_1t^2=x_{02}+v_{02}t+\dfrac{1}{2}a_2t^2,$$ $$(a_1-a_2)t^2+2(v_{01}-v_{02})t+2(x_{01}-x_{02})=0.$$

Substituting the numbers into the equation, we get:

$$(4.5-0)t^2+2\cdot(0-14)t+2\cdot(0-49)=0,$$ $$4.5t^2-28t-98=0.$$

This quadratic equation has two roots: $t_1=8.72\ s$ and $t_2=-2.5\ s.$ Since time can't be negative the correct answer is $t=8.72\ s.$

Therefore, the first player catches his opponent after 8.72 seconds starting from rest.

(b) We can find the distance that the first player traveled in that time from the kinematic equation for the first player:

$$x_1=x_{01}+v_{01}t+\dfrac{1}{2}a_1t^2,$$ $$x_1=0\ m+0\ \dfrac{m}{s}\cdot8.72\ s+\dfrac{1}{2}\cdot4.5\ \dfrac{m}{s^2}\cdot(8.72\ s)^2=171\ m.$$