# magnitude of acceleration

• February 19th 2008, 04:11 AM
galactus
magnitude of acceleration
Hello All:

Here's a problem I am ashamed to admit has me stymied. I am sure someone here has an idea.

"Professor lives S miles from the university and it takes him T minutes to drive from home to work. Prove that at some instant the magnitude of the acceleration of his car is at least $\frac{4S}{T^{2}}$.

I think we can use the MVT.

Wouldn't he be accelerating to the midpoint and then decelerating?.

I was thinking perhaps the max acceleration would be at S/4.
• February 19th 2008, 07:58 AM
topsquark
Quote:

Originally Posted by galactus
Hello All:

Here's a problem I am ashamed to admit has me stymied. I am sure someone here has an idea.

"Professor lives S miles from the university and it takes him T minutes to drive from home to work. Prove that at some instant the magnitude of the acceleration of his car is at least $\frac{4S}{T^{2}}$.

I think we can use the MVT.

Wouldn't he be accelerating to the midpoint and then decelerating?.

I was thinking perhaps the max acceleration would be at S/4.

Assuming that the professor ends his/her trip at 0 m/s. (I suppose he could crash his car at the end of the problem, but then we can calculate the deceleration at the end and find out it is enormously larger than the given answer. :) ) So the least possible acceleration will be given by a constant acceleration to the half-way point and the same magnitude constant deceleration to the destination. (That this gives the "minimum possible maximum" acceleration is guaranteed by the MVT.)

So the acceleration may be calcuated by
$d = \frac{1}{2}at^2$

The professor travels S/2 m to get to the half-way point, in a time T/2.

Thus
$\frac{S}{2} = \frac{1}{2}a \left ( \frac{T}{2} \right )^2$

Solve for a. This is the least possible maximum acceleration to get to the destination on time.

-Dan
• February 19th 2008, 08:04 AM
wingless
Quote:

Originally Posted by galactus
Wouldn't he be accelerating to the midpoint and then decelerating?.

You're right about that. As we're looking for the least acceleration, we will take the movement uniformly increasing and decreasing. (You can easily prove this)

Here's the V-t graph:

http://img513.imageshack.us/img513/6215/gphza6.png

Area under the V-t graph gives the distance travelled.
Then, the area from 0 to T under the graph is S.

$Area = \frac{1}{2} T \cdot V_{max} = S$

$S = \frac{T\cdot V_{max}}{2}$

$V_max = \frac{2S}{T}$

Slope of the lines (which also means the derivative) gives the acceleration.

$Slope = a = \frac{\Delta y}{\Delta x} = \frac{V_{max}-0}{\frac{T}{2}-0}$

$a = \frac{2\cdot V_{max}}{T}$

Plug the $V_{max}$ we found before,
$a = \frac{2\cdot \frac{2S}{T}}{T}$

$a = \frac{4S}{T^2}$

Edit: topsquark was faster :P
• February 19th 2008, 08:42 AM
galactus
I knew I'd feel stupid. I forgot all about the $\frac{1}{2}at^{2}$

I was also trying to use the MVT which wasn't necessary.

Thanks.
• February 19th 2008, 07:19 PM
topsquark
Quote:

Originally Posted by galactus
I knew I'd feel stupid. I forgot all about the $\frac{1}{2}at^{2}$

I was also trying to use the MVT which wasn't necessary.

Thanks.

The MVT is needed to prove that this is the minimum necessary acceleration, so I wouldn't say that it wasn't necessary. I'd say it's rather critical. (Though it admittedly doesn't help the actual calculation.)

-Dan
• February 20th 2008, 04:33 AM
galactus
That's pretty close to what wingless showed, isn't it?.
• February 20th 2008, 07:51 AM
topsquark
Quote:

Originally Posted by galactus
That's pretty close to what wingless showed, isn't it?.

Yeah, I guess you are right about that. (Nod)

-Dan