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April 16, 2014

April 16, 2014

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

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 4:58pmsorry the slash after the three should actually be a period, the equation is only 24t/2t+3

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

- calculus-follow up for damon -
**Kenneth**, Saturday, November 22, 2008 at 5:48pmthank 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-follow up for damon -
- calculus -
**Damon**, Saturday, November 22, 2008 at 5:48pmNo!!!

acceleration = dv/dt

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

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