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
$$\left(\frac{ds}{dt}\right)^2$$!