It is given that acceleration is constant, so can we infer that average speed and velocity are the same?? Moreover, circular motion is out of the question, as the function of x(t) where x=displacement, suggests, that for any t>=0, displacement can not be zero...

This is the conceptual problem I am facing in a question: My teacher was reading out the question, and it was asked only to find the avg velocity from the acceleration. She, on her own, added a part to it, asking us to also find avg speed, and then, while discussing solutions, said that a graph must be made in order to solve this...so do you think that it is absolutely necessary? Moreover, if my premise is flawed, then how can the graph even help??

Thanks in advance!

  • $\begingroup$ velocity is a vector, the same as position (x,y,z) is a vector. So is acceleration a vector. speed is the square root of the dot product of velocity with itself. You should use the tag "homework" $\endgroup$ – anna v Apr 30 '13 at 10:04
  • $\begingroup$ that is understood, however, I fail to see how this answers my question $\endgroup$ – Saurabh Raje Apr 30 '13 at 10:24
  • $\begingroup$ average speed and velocity are not the same. speed is the measure of velocity.this is a comment, not an answer. look at en.wikipedia.org/wiki/Circular_motion $\endgroup$ – anna v Apr 30 '13 at 10:51
  • $\begingroup$ That is a valid point. I forgot to mention that we were talking in terms of just magnitude, and to make matters more clear, the actual question was: x(t) = (t+9)^2 and the time interval can be anything you want (I dont remember it exactly, nor do I think it makes a difference.. $\endgroup$ – Saurabh Raje Apr 30 '13 at 10:55

Let's see with help of an example.

Let the particle is at $(0,0)$ moving with speed of $2m/s$ at $t=0$ and is subjected to acceleration $-2\hat i\ \text{m/s}^{-2}$.Now see after $2 seconds$.

We see displacement is zero,and distance travelled =$2m$ .Also acceleration is constant but still $\langle speed\rangle\not=0$ whereas $\langle velocity \rangle=0$

Now let's see the graphs for this scenario. enter image description here

You see distance traveled is area under v-t curve (all positive) but displacement is are with signs. The area under X-axis is negative.

So, graphs can help in cases where velocity changes direction.

  • $\begingroup$ did you mean moving with velocity = 2m/s @t=0? $\endgroup$ – Saurabh Raje Apr 30 '13 at 11:02
  • $\begingroup$ Moreover, in this case, the displacement is zero with positive time. When x(t) = (t+9)^2, displacement can never be zero for a positive time, right? $\endgroup$ – Saurabh Raje Apr 30 '13 at 11:05
  • $\begingroup$ @SaurabhRaje Yes, with zero initial velocity such cases are not possible. $\endgroup$ – ABC Apr 30 '13 at 11:07
  • $\begingroup$ so, returning to my question, do you think tht avg velocity and avg speed have same magnitude? $\endgroup$ – Saurabh Raje Apr 30 '13 at 11:08
  • $\begingroup$ No. Not always. The cases in which the velocity changes direction(goes zero in motion) have different values for $\langle speed \rangle$ and $\langle velocity \rangle$ $\endgroup$ – ABC Apr 30 '13 at 11:12

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