Question

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 4.0 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 0.14 m/s2, determine each of the following:
a. How long does it take him to catch his opponent?
b. How far has he traveled in that time?

Answer 1

Answer:

a)$t=59.817652833125294s \approx 59.8s$

b)$D_1=894.01m$

Explanation:

From the question we are told that

Speed of opposing player $V_2=4.0m/s$

First player chase his opponent after$t=2.80t$

Acceleration of  first player $a=0.14 m/s2$

Let time of catch be  $t_c$

a)Generally the Equation for distance covered is mathematically given as follows

Distance to catch First opponent

$D_1=\frac{1}{2}(0.14)t^2$

$D_1=0.07t^2$

Distance to covered Second opponent

$D_2=(2.8+t)*4$

Generally when first opponent catch the second opponent it is represented mathematically as

$D_2=D_1$

$0.07t^2=(2.8+t)*4$

$0.07t^2=11.2+4t$

$0.07t^2-4t-11.2$

$t=59.817652833125294s \approx 59.8s$

b)Generally the the total time traveled by the first opponent is mathematically given as

$D_1=\frac{t^2}{4}$

$D_1=\frac{59.8^2}{4}$

$D_1=894.01m$