Ear the end of a marathon race, the first two runners are separated by a distance of 45.0 m. the front runner has a velocity of 3.45 m/s, and the second a velocity of 4.25 m/s. (a) what is the velocity (in m/s) of the second runner relative to the first? m/s faster than the front runner (b) if the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity? the first runner will win. the second runner will win. (c) what distance (in m) ahead will the winner be when she crosses the finish line? m

At the time t = 0, the distance is 45m

the front runer has a velocity of 3.45 m/s, the second one has a velocity of 4.25 m/s

then the position of both runes, puting the zero of the position where the runer 1 is:

p1(t) = (3.45m/s)*t

p2(t)= (4.25m/s)*t - 45m

a) relative velocity: this is the differenceof both velocities: v2 - v1 = 4.25m/s - 3.45m/s = 0.8m/s

this means that the first runner sees the second one with a positive velocity.

b) if the first runner is 250m away from the finish line, then:

the first runner needs:

p1(t) = 250m = (3.45m/s)*t

t = (250/3.45) s = 72.5 seconds

for the runner 2

p2(t) = 250m = (4.25m/s)*t - 45m

250m + 45m = 295m = (4.25m/s)*t

(295/4.25) s = t = 69.4 seconds

So the runner 2 needs less time, this means that he will reach the finish line first

c) we know that at t= 69.4 seconds, the second runner is in the finish line, we can imput that thime in the position of the runner 1:

p1(69.4s) = (3.45m/s)*69.4s = 239.5m

and at this time, the runner 2 is already at the finish line that was 250m away from the first runner, so the distance is:

250m - 239.5m = 10.5m

true is the correct answer

hope i

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west