Posted by **Kenneth** on Saturday, November 22, 2008 at 4:57pm.

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**, Saturday, November 22, 2008 at 4:58pm
sorry the slash after the three should actually be a period, the equation is only 24t/2t+3

- calculus -
**Damon**, Saturday, November 22, 2008 at 5:02pm
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**, Saturday, November 22, 2008 at 5:08pm
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**, Saturday, November 22, 2008 at 5:40pm
thank you so much! as a follow up question, for acceleration of the two runners, is acceleration the anti derivative of velocity???

- calculus -
**Damon**, Saturday, November 22, 2008 at 5:48pm
No!!!

acceleration = dv/dt

- calculus -
**Damon**, Saturday, November 22, 2008 at 5:55pm
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**, Saturday, November 22, 2008 at 5:56pm
By the way, position is the anti-derivative of velocity

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