Posted by Kenneth on .
Two runners, A and B run on a straight race track for 0</=t</=10 seconds. The graph below, which consists of two line segments, shows the velocity, in meters per second, of runner A. The velocity of runner B, in meters per second, is given by the function v defined by v(t)=24t/2t+3/
Although I can't post the graph, there are two points labeled, (3,10) and (10,10).
ind the velocity of runner A and runner B at t=2 seconds. I know you can't see the graph, but is there any way you could guide me on how to do this problem?? thank you

calculus 
Kenneth,
sorry the slash after the three should actually be a period, the equation is only 24t/2t+3

calculus 
Damon,
v(t)=24t/(2t+3) [[ Maybe ?????? ]]
Although I can't post the graph, there are two points labeled, (3,10) and (10,10).
[[ and (0,0) ]] maybe???? 
calculus 
Damon,
MAKING THOSE ASSUMPTIONS ABOUT WHAT YOU WERE GIVEN!!!!
At two seconds for runner A
I think runner A went from (0,0) to (3,10) in a straight line
then:
y = m x + b
(0,0) > b = 0
m = 10/3
so
y = (10/3)x
for x = 2 y = 20/3 or 6.67
Runner B if y = 24 x / (2x+3)
where x = 2
y = 48 / (4+3) = 48/7 = 6.86 
calculus 
Kenneth,
thank you so much! as a follow up question, for acceleration of the two runners, is acceleration the anti derivative of velocity???

calculusfollow up for damon 
Kenneth,
thank you so much or your help! as a follow up question, for acceleration of the two runners, is acceleration the anti derivative of velocity???

calculus 
Damon,
No!!!
acceleration = dv/dt 
calculus 
Damon,
dv/dt for A = 10/3 (constant slope until t = 3)
dv/dt for B
v(t)=24t/(2t+3)
a = dv/dt = [(2t+3)(24) 24t(2) ]/(2t+3)^2
put in t = 2
a(2) = dv/dt = [ 7(24) 96 ] /(49)
= 72/49 = 1.47 
calculus 
Damon,
By the way, position is the antiderivative of velocity