March 2, 2015

Homework Help: AP Physics

Posted by AP Physics on Saturday, June 13, 2009 at 1:12pm.

can you please explain this to me...

Were given this eqaution

x = At^2 + B

were A = 2.10 m/s^2
B = 2.80 m

ok and this is an example problem and it's wlking us through it

find the instantaneous velocity at t=t2=5.00s equlas the slope of the tangent to the curve at point P2 whoen in Fig. 2-12b and we could measure this slope off the gropah to obtain v2. we can calculate v more precisely, and for any time using the give formula

x = At^2 + B,

which is the engiene's position x at time t. Using the calculus formula for derivitves (THIS STEP I'M LOST)

(d/(dt))(Ct^n)= nCt^(n-1) and (dC)/(dt) = 0,

where C is any given constant, then

v = (dx)/(dt) = (d/(dt))(At^2 + B) = 2At.

We are given A = 2.10 m/s^2, so for t=t2=5.00s,

v2 = 2At = 2(2.10 m/s^2)5.00s = 21.0 m/s

ok it has been a long time sense I have taken calculus. I understand why instanteous velociy is equal to (dx)/(dt) but I do not understand this problem. Can you please explain it to me. I got lost were I wrote (THIS STEP I'M LOST)

I don't understand this
"(d/(dt))(Ct^n)= nCt^(n-1) and (dC)/(dt) = 0,

where C is any given constant, then

v = (dx)/(dt) = (d/(dt))(At^2 + B) = 2At."

Please explain to me the constant thing and what d is all by itself and what is dC and why they divide it by dt and how they got 2At

Thank you!

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