Solving an Initial Value Problem for Acceleration: Finding s at t = 1 sec

[Sedm]

The problem:

The acceleration of a particle moving back and forth on a line is a = (d^2)s/d(t^2) = (pi)^2 cos(pi)(t) m/sec^2 for all t. If s = 0 and v = 8 m/sec when t = 0, find s when t = 1 sec.

My work:

(d^2)s/d(t^2) = (pi)^2 cos(pi)(t)

ds/dt = (pi)^2 sin(pi)(1) + C

ds/dt = (pi)^2(0) + C

0 = C

Then..

ds/dt = (pi)^2 sin(pi)(t)

s = -(pi)^2 cos (pi) (8)

0 = -(pi)^2(1) + C

(pi)^2 = C

So my answer turned out to be (pi)^2 meters. I'm not so sure that that's the correct answer though.

Any help is appreciated.

 

[Gale]

well, oook.

First of all, you start with acceleration as a function of time yes? a(t) = \pi^2 cos(\pi t) when you integrate \frac{d^2 s}{dt^2}=\pi^2 cos(\pi t) this is an equation for velocity as a function of time v(t)= \frac {ds}{dt} right?

so, first when you integrate, cos(\pi t) dt its the opposite of the chain rule, so you have to divide by the coefficient of t, in this case, pi. so you get,
v(t) = \frac{ds}{dt}=\pi sin(\pi t) +C

then to solve for C you plug in the given values for v and t and solve.
Then because you want s(t), you integrate once more. and you solve for C again with the given initial values of s and t. then you plug in t=1 into your final equation and solve for s.

 

[Hurkyl]

A quick notational fix for the original poster:

The notation:

\frac{d^2 s}{dt^2}

means

\frac{d^2 s}{(dt)^2}

or equivalently,

\left( \frac{d}{dt} \right)^2 s

 

[Gale]

oh man, you scared me when i saw your post after mine... i thought i'd done something wrong! whew.

 

[Hurkyl]

Nah, your work looks fine! The only thing I might consider complaining about is that you did virtually all the work for the OP.

 

[Sedm]

Ah, I think I've got it now. I just hope I won't confuse this with anything else on my test tomorrow. :P

Thanks!

 

[HallsofIvy]

Sedm: another notational point. Many of us might be inclined to read
cos(pi)(t) as {cos(pi)} t, in which case, your original calculation would be correct: the derivative of that would be {cos(pi)} but NOT the anti-derivative! Clearly you meant cos(pi t). The anti-derivative of that is
(1/pi) sin(pi t). I don't know why had "1" in place of t.

Hurkyl: ?? What??
\frac{d^2 s}{dt^2} is the second derivative. It definitely is not
\frac{d^2 s}{(dt)^2}
(I'm not even sure what that could mean!)
Yes, you could write that as
\left( \frac{d}{dt} \right)^2 s
but you had better make it clear that that is NOT
\left(\frac{ds}{dt}\right)^2!