# Uniform acceleration lab

1. Oct 21, 2005

### DB

lab question on uniform accelerated motion

7) often a velocity time graph does not pass through the origin. suggest some reasons why this may happen.

sure enough my velocity vs. time graph (of average velocities calculated from the slope of tangent lines on the midpoint of a time interval) didnt pass through the origin.

so id like to know if im right on this:
the reason why a "v vs t" doesnt always pass through the origin is because you cant instantaneously reach a constant acceleration, a cart for example, will have to accelerate un-uniformely until it reaches a constant acceleration.

that make sense?

2. Oct 21, 2005

### whozum

If it doesnt pass through the origin it means that there is an initial velocity, remember that the equation of a line can be given as

[tex] y = mx + b [/itex] with m the slope and b the y intercept, this relates directly to

[tex] v = at + v_0 [/itex] which is the same equation except with different letters (that you know the applied meaning of). What is the equivalent of the y intercept in this equation? $v_0$, the initial velocity.

3. Oct 21, 2005

### lightgrav

>>you cant instantaneously reach a constant acceleration

that's not likely to be an issue.

It takes time to accumulate velocity from an acceleration,
but unless you released the cart "gradually" on purpose,
the acceleration should reach its value within a few millisec.

Although you didn't really describe your lab setup,
I would guess that the cart was moving before it got to
where you started timing from.

Did you find average velocity at the "location mid-point"
of the time interval, not the "time mid-point"?

Did your carts have reduced acceleration at higher speed?

Last edited: Oct 21, 2005
4. Oct 22, 2005

### daniel_i_l

>>you cant instantaneously reach a constant acceleration

If you drop a ball dosn't it instantaneously reach an acceleration of
9.8m/s^2?

maybe you mean that you cant instantaneously reach a constant speed

5. Oct 23, 2005

### DB

does it?
no, i didnt mean this.

6. Oct 28, 2011

### drezden9

If you push a ball off a tower, the moment that normal force is no longer neutralizing the weight of the ball, f=ma so the net force with respect to y will have to have some acceleration value. So yes, if you drop a ball it does reach an acceleration of 9.8 m/s^2 rather instantaneously.

If you velocity vs time graph doesn't pass through the origin, it just means that at t=0, there was some velocity value. Yes, there does need to be some force that accelerates the object to that initial velocity, but assuming the force's presence is not constant, that initial acceleration is negligible. What will cause uniform acceleration is a force that is ever-present, like gravity during free-fall or combustion during an explosion.