- #1

- 1,067

- 92

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- B
- Thread starter rudransh verma
- Start date

If the constant acceleration was constant in the past, then the average velocity would be attained as an instantaneous velocity somewhere within the time interval.

- #1

- 1,067

- 92

Physics news on Phys.org

- #2

Mentor

- 8,345

- 5,345

$$

s = (u + \frac{1}{2} a t) t

$$

where the parenthesis represents the velocity at time ##t##. You will get negative displacements whenever that velocity is negative.

- #3

Science Advisor

Homework Helper

- 12,227

- 6,918

You can get a negative displacement if ##u## (the initial velocity) is negative.rudransh verma said:

You can get a negative displacement if ##a## (the ongoing acceleration) is negative and you wait long enough (or look far enough into the past).

You can get a negative displacement if ##t## is negative and ##u## is positive. So you are looking at displacement at some time in the past.

Graphically, you are looking at a parabola for which some portion extends below the t axis.

The way I read the second equation of motion is that ##s = ut## (Current displacement is equal to initial velocity times elapsed time). But if there is a constant acceleration, ##a##, then that results in an additional displacement of ##\frac{1}{2}at^2## that simply adds on.

As I understand @DrClaude, he takes the viewpoint that for constant acceleration, the average velocity is attained halfway through the time interval, so it is given by ##u + \frac{1}{2}at##. Multiply average velocity by the total time interval ##t## to get ##s=(u + \frac{1}{2}at)t##.

Six of one, half dozen of the other.

Last edited:

- #4

- 1,067

- 92

So ##u+\frac12at## is average velocity and when this is -ve then s is also -ve. The eqn of ##s=ut+\frac12at^2## is same as ##{x_2}-{x_1}= {v_{avg}}t##.jbriggs444 said:As I understand @DrClaude, he takes the viewpoint that for constant acceleration, the average velocity is attained halfway through the time interval, so it is given by u+12at. Multiply average velocity by the total time interval t to get s=(u+12at)t.

u is the initial velocity and due to a acceleration somewhere in time interval t ##{v_{avg}}## is achieved. v is the final velocity not included in the eqn ##u+\frac12at##

Last edited:

- #5

Science Advisor

Homework Helper

- 12,227

- 6,918

Yes -- if we take ##t## as positive. But the equation still holds for ##t## negative if the constant acceleration was constant in the past.rudransh verma said:So ##u+\frac12at## is average velocity and when this is -ve then s is also -ve.

Here I am not sure whether there is a language difficulty or whether you are contemplating a non-constant [1 dimensional] velocity and invoking the mean value theorem. Yes, under these conditions, I believe that the mean value theorem guarantees that the average velocity be attained as an instantaneous velocity somewhere within the time interval.rudransh verma said:u is the initial velocity and due to a acceleration somewhere in time interval t ##{v_{avg}}## is achieved.

We normally restrict ourselves to constant accelerations when using the SUVAT equations so that instantaneous velocity is equal to average velocity exactly at the midpoint of the interval.

- #6

- 1,067

- 92

Constant acceleration. Yes!jbriggs444 said:Here I am not sure whether there is a language difficulty or whether you are contemplating a non-constant but continuous acceleration and invoking the mean value theorem.

Share:

- Replies
- 1

- Views
- 96

- Replies
- 2

- Views
- 633

- Replies
- 7

- Views
- 1K

- Replies
- 1

- Views
- 475

- Replies
- 4

- Views
- 559

- Replies
- 7

- Views
- 841

- Replies
- 3

- Views
- 684

- Replies
- 9

- Views
- 739

- Replies
- 8

- Views
- 806

- Replies
- 21

- Views
- 953