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Why is speed and acceleration negative when $V_1$ of an object is say 150m/s, $V_2$ is 0 m/s and $\Delta d=0.50\,\rm m$? I found the time it takes which is 0.0033s and the acceleration to be 90909.09 m/s$^2$.

My questions:

  1. What do negative velocity and negative acceleration mean?
  2. Is it negative because it is slowing down?
  3. Why is the acceleration so big? Is that even reasonable?
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  • $\begingroup$ Without using vectors, such problems tend to assume a single spatial dimension. Negative quantities simply mean 'opposite in direction to positive' in this context. $\endgroup$ – Xeren Narcy May 28 '15 at 2:04
  • $\begingroup$ would that mean the object is moving backwards ? or slowing down ? $\endgroup$ – Bethany May 28 '15 at 2:08
  • $\begingroup$ I'm pretty sure that your answer for acceleration is wrong. $\endgroup$ – Jimmy360 May 28 '15 at 2:16
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Remember that velocity and acceleration are vectors, they have a magnitude and direction. Speed, on the other hand, is a scalar quantity which only has magnitude.

If you are moving (in a car) towards the right and apply the breaks, you are decelerating (slowing down)--your acceleration is directed towards the left while your velocity ("speed" in the diagram) is pointed towards the right (and is getting smaller):

enter image description here
(source)

So yes, the negative sign indicates a slowing down. A negative velocity would mean that the object is now moving in the opposite direction it was initially (or what you call "positive" direction, we could have made left the positive direction in the previous example).

The magnitude of your (negative) acceleration seems pretty high (though reasonable order of magnitude as to what I get below) given the high rate of speed initially ($v_1=150\,\rm m/s$), the zero final speed, and the extremely short stopping distance ($d=0.5\,\rm m$). You can use one of your kinematics equations, $$ v_2^2=v_1^2+2a\Delta x, $$ to get that $a=-22,500\,\rm m/s^2$, which is about 1/4 of what you have. Not sure where your mistake is, but I suspect it's in the calculation of the time (which doesn't seem necessary to me).

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  • $\begingroup$ if i used the acceleration i got and plugged it into fnet=ma ti find the force the object would hit another object at it would be negative. what would this mean? because i calculated acceleration so to get acceleration i used a= Δv/Δt so Δv= 0-150 = -150 so a=-150/0.0033s = -45454.54 m/s^2. when i plug that into fnet = ma, there is really only one force so Fa = ma and ma = (55)(-45454.54) i get Fa = -2 5000 000. what does that mean ? is the first object losing that much force ? $\endgroup$ – Bethany May 28 '15 at 2:25
  • $\begingroup$ The problem is you are viewing it as losing force and that's not at all the case. The negative sign associates a direction to the force, that's it. $\endgroup$ – Kyle Kanos May 28 '15 at 2:28
  • $\begingroup$ so its is force put in the backwards direction? $\endgroup$ – Bethany May 28 '15 at 2:39
  • $\begingroup$ That is correct. $\endgroup$ – Kyle Kanos May 28 '15 at 2:46
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enter image description here

In this acceleration graph, velocity is going down. Remember that in this graph, the slope is acceleration, so the acceleration is negative. Negative acceleration is what some people call acceleration (slowing down). If the velocity starts negative, however, then negative acceleration is speeding up.

For negative velocity, imagine a number line.

enter image description here

Imagine that you go to 1 in one second. Your velocity is 1 m/s. Now imagine going back to 0 from 1 in 1 second. Your velocity is -1 m/s.

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    $\begingroup$ Negative acceleration would be equivalent to slowing ONLY if the velocity is positive. If the velocity is zero or negative, a negative acceleration would actually be equivalent to the object moving faster. $\endgroup$ – Bill N May 28 '15 at 2:25

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