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

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

No!!!

acceleration = dv/dt

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

By the way, position is the anti-derivative of velocity

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

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????

thank you so much! as a follow up question, for acceleration of the two runners, is acceleration the anti derivative of velocity???

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???

Certainly! Even without the graph, we can find the velocities of runner A and runner B at t=2 seconds using the provided information.

To find the velocity of runner A at t=2 seconds, we need to look for the line segment on the graph that starts at t=0 and ends at t=10. Since the second point labeled on the graph is (10,10), we can assume that the velocity of runner A remains constant at 10 meters per second for the entire duration of the race. Therefore, the velocity of runner A at t=2 seconds is also 10 meters per second.

To find the velocity of runner B at t=2 seconds, we can use the function v(t) = 24t / (2t + 3). Simply substitute t=2 into the equation:

v(2) = 24(2) / (2(2) + 3)
= 48 / (4 + 3)
= 48 / 7

So, the velocity of runner B at t=2 seconds is approximately 6.86 meters per second.

Remember, in this scenario, runner A has a constant velocity of 10 meters per second, while the velocity of runner B changes according to the given function.

I hope this helps! Let me know if you have any further questions.