# Acceleration

• Aug 2nd 2006, 09:01 AM
bret80
Acceleration
If an object moves according to a function of time that is defined by the equation:

y= -4t^3 + 20t^2 + 80t + 100.

What is the acceleration when time t=1?
(The acceleration is the rate of change of velocity with respect to time, that is, a= d^2y/dt^2)
• Aug 2nd 2006, 09:21 AM
ThePerfectHacker
Quote:

Originally Posted by bret80
If an object moves according to a function of time that is defined by the equation:

y= -4t^3 + 20t^2 + 80t + 100.

What is the acceleration when time t=1?
(The acceleration is the rate of change of velocity with respect to time, that is, a= d^2y/dt^2)

In America acceleration is the second derivative of distance.

Thus,
$\displaystyle y'=-12t^2+40t+80$
$\displaystyle y''=-24t+40$
Evaluate function at $\displaystyle t=1$,
$\displaystyle y''(1)=-24(1)+40=16 \frac{\mbox{units}}{\mbox{sec}^2}$
• Aug 2nd 2006, 10:13 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
In America acceleration is the second derivative of distance.

Actually bret80 is correct. Unless otherwise stated the term acceleration refers to a vector, rather than a scalar. Acceleration is always the second time derivative of the displacement function and first time derivative of the velocity function. Unfortunately even professors get lazy in their terminology and students tend to forget the difference. In fact, I've even found some students (not mine!) at the end of the semester who don't recall that the term displacement was ever used in the course! :(

-Dan
• Aug 2nd 2006, 10:24 AM
ThePerfectHacker
The way I understand it is:
Displacement (Is a vector function to assigns position as well). Its derivative is Velocity (a vector function that assigns the direction as well).

Distance (Is a real function that assigns a number which corresponds to the length traveled). Its derivative is speed (a real function that assigns a number which correspond to only its current speed, nothing more).

However, I do not know how to think of acceleration. Is it the derivative of a real or vector function?
• Aug 2nd 2006, 10:34 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
The way I understand it is:
Displacement (Is a vector function to assigns position as well). Its derivative is Velocity (a vector function that assigns the direction as well).

Distance (Is a real function that assigns a number which corresponds to the length traveled). Its derivative is speed (a real function that assigns a number which correspond to only its current speed, nothing more).

However, I do not know how to think of acceleration. Is it the derivative of a real or vector function?

By definition, acceleration is the derivative of velocity...ie. it is a vector function.

However, just to confuse the issue it is often convenient when doing problems in one dimension to "drop" the vector from the acceleration. This is why, when asked what "g" is, people often answer with "9.8 m/s^2" rather than the more correct "9.8 m/s^2 toward the center of the Earth" (or more simply "downward.") Really we should always be including a unit vector in the direction of the + coordinate when reporting an acceleration in a 1-D problem. To further confuse the issue is the term "deceleration," which is nothing more than an acceleration in a direction opposite to the velocity, which students often confuse with either always being negative (ie. in a direction opposite the unit vector) or decreasing in value.

This is why professors often lose the vector nature of the acceleration. (And usually don't bother to correct it, since it doesn't "need" to be corrected until the student gets into Advanced (undergrad) Mechanics.) A similar bad habit goes for confusing velocity and speed. I personally think it's a lazy habit and bad for the students.

-Dan