A hockey player is standing on his skates on a frozenpond when an opposing playe

Question

A hockey player is standing on his skates on a frozenpond when an opposing player, moving with a uniform speedof 10 m/s, skates by with thepuck. After 4.0 s, the firstplayer makes up his mind to chase his opponent. If he acceleratesuniformly at 5.0 m/s2, determine each of thefollowing. (a) How long does it take him to catch his opponent? (Assume theplayer with the puck remains in motion at constantspeed.)
(a) How long does it take him to catch his opponent? (Assume theplayer with the puck remains in motion at constantspeed.)

Explanation / Answer

Hi. If one player catches the other, it means they travelled the samedistance. At t=0, the player with the puck is at a distance d = (10 m/s)( 4.0s) = 40 m, while the other player is still at d = 0. So, over time, the player with the puck will be at a distance d= (40 m) + v t from the starting point. During that same time, the chasing player will be at d' = (1/2) at2 . They meet when d = d' : (40 m) + v t = (1/2) a t2      a t2 - 2 v t - 2(40 m) = 0. To find t, we need to solve that quadratic equation for thevariable t: t = answers of the form " [ -b ± (b2-4ac) ] / 2a " = [ 2v ± ((-2v)2- 4a(-80) ) ] / 2a =   ...alittle algebra... = [ v ± ( v2 + 80a) ] /a Replace the velocity v and acceleration a by their values ( v= 10 m/s , a = 5.0 m/s2) And try not to make any mistakes when you "plug in" thenumbers.... You will get two solutions (because of the ±). Onlyone of those will make sense. This value of t is the answer youwere looking for.

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