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Homework Help: Quick question

  1. Jul 27, 2013 #1
    Hi guys, I've been having a pretty hard time understanding exactly what acceleration is, the definitions I usually see are always different, so from what I've read, I know acceleration is a vector quantity and the definition is basically change in velocity?
    So if lets say a car is going 45 km/h east, and it changes its route to and speed (magnitude) to 35 km/h west, that would be change in acceleration right? Or do you have to account in the time as well, acceleration = velocity / time?
    I may be over thinking this concept, but would be nice if someone can really clarify this for me, thanks again, huge fan of the physics forums.
  2. jcsd
  3. Jul 27, 2013 #2
    No. It's change in velocity over a change in time. :wink:

    That would be a change in velocity. Not a change in acceleration.

    What do you know about calculus? It's easy to explain if you know derivatives.
  4. Jul 27, 2013 #3
    Ah right change in velocity over time lol. I've taken calculus one, so yes I know about derivatives and such :p.
  5. Jul 27, 2013 #4
    This is essentially correct. To make it fully correct, say the rate of change in velocity. Rate implies that the "change" is divided by the time over which the "change" happens.

    Just like velocity itself is the rate of change in displacement.

    Mathematically, these "rate of change" things are known as "derivatives". Velocity is the derivative of displacement, and acceleration is the derivative of velocity (and the second derivative of displacement).

    Change in velocity. It is -80 km/h. Note the minus sign, I assume that the eastern direction is positive.

    Not exactly. You do not need just time, you need the time it took for the change to occur. In your example, you did not give that time. If it took the car 20 seconds to undergo that change (say, by going through a roundabout), then the rate of change in velocity is (80 km/h) / (20 s) = 1.1 m/s2.
  6. Jul 27, 2013 #5
    Alright. That's good. Then, I'll introduce you to a new concept.

    Since we can write a vector, let's say ##\vec{x}=\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}##, as a sum of the form ##\vec{x}=x_1\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}+x_2\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}+x_3\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}##, we can say that vectors are differentiated by components. That is, ##\frac{d\vec{x}}{dt}=\frac{d}{dt}\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}=\begin{bmatrix}\frac{dx_1}{dt} \\ \frac{dx_2}{dt} \\ \frac{dx_3}{dt}\end{bmatrix}##. This is the definition of the derivative of a vector (in Euclidean space) with respect to a scalar.

    Thus, if we call ##\vec{x}## our displacement vector, then ##\vec{v}=\frac{d\vec{x}}{dt}## is the velocity vector and ##\vec{a}=\frac{d^2\vec{x}}{dt^2}## is the acceleration vector.
  7. Jul 27, 2013 #6
    Thanks Voko, I remember how the concept of derivatives relates to acceleration and other physics quantities now, your definitions were very helpful.

    I see somewhat of what you're saying, but at other places I feel lost lol, I'll have to fresh up on my linear algebra I think to fully understand the acceleration vector as you presented it. It is very much appreciated though, thank you!

    Thanks guys!
  8. Jul 28, 2013 #7
    I'll put it in layman's terms.

    You're travelling on a cycle.
    You're speed after 1 second in 5 seconds is
    0 second- 0 m/s
    1 second- 2 m/s
    2 seconds - 4 m/s
    3 seconds - 6 m/s
    4 seconds -8 m/s
    5 seconds - 10 m/s

    So what is happening is, you are increasing your speed by 2 m/s after every second. This makes your acceleration 2m/s/s or 2m/s^s.

    This is just linear acceleration. For direction change, vector diagrams aer necessary.
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