Kinematics equations for constant acceleration

[kingwinner]

1)

This is one of the 1D kinemantic equation:

 

displacement=(initial velocity)(delta t) + (1/2)(acceleration)(delta t)^2

 

and is solving for delta t

 

======================

 

I don't understand the circled parts! Why can you just omit the negative v2 and take the positive v2? And what is the reason of doing this?

 

(Note that this finally derives back to the equation acceleration = change in velocity / time)

 

If someone do know the answer, please help me! Thank you!

[CraigD]

Consider the equation

displacement=(initial velocity)(delta t) + (1/2)(acceleration)(delta t)^2

Which solves, via the quadratic formula, to

delta t = (initial velocity) +/- ( (initial velocity)^2 +2(acceleration)(displacement))^.5 )/(acceleration)

 

With some example values

displacement= -15 m

initial velocity= 10 m/s

acceleration= -10 m/s/s

then

delta t = (10 + (100 +300)^.5)/10 = 3 sec

and t = (10 - (100 +300)^.5)/10 = -1 sec

 

What does this mean?

 

It would mean that, after 3 seconds, an object that is 15 meters high, moving upward at 10 m/s, will strike the ground, and, 1 second before, it was also at ground level. In other words, the equation could describes an object thrown upward from the ground at 20 m/s, with a “snapshot” of its height and velocity taken 1 second later. 3 seconds after the snapshot, the ball falls back to the ground.

[Qfwfq]

I think Craig that his problem is specificaly the ± sign, and its disappearance.

 

King if you ask things a bit more reasonably, as well as nicely, you might get help more easily. ;)

 

The ± is due to solving the second degree equations but, in this case, one of the two solutions is negative and is only of interest if you want to extrapolate backward in time. This makes sense if the force has been the same until now, it doesn't if you only consider the motion from now on.

 

HTH