Acceleration Problems (1 Viewer)

[Dt 08]

Hey


I am having trouble with acceleration formulae

My first problem concerns the formulae a=(u-v)/t , a=(delta)v/(delta)/t and all the other formulae related to these such as r=0.5(2u+at)t

Are these formulae only applicable in constant acceleration?? Also, how do we find out the acceleration in non-uniform acceleration. Is da formula such as a=(delta)v/(delta)/t still applicable?



Also, how is the formula for constant velocity (u+v)/2 and constant acceleration (??) derived?




Thanking you in advance

 

[alcalder]

This is from a previous post I did:

OK to do it the Ext 1 way, first take the initial velocity v at an angle θ to the horizontal and resolve it into vertical (y-direction) and horizontal (x-direction) components.

INTIAL CONDITIONS
at t=0
u_y = vsinθ
u_x = vcosθ

If you start on the horizontal, then at t=0, x=0 AND y=0
If you start on a cliff, say 100m off the ground, at t=0, x=0 and y=100

Up is always + and down is - direction.

DERIVING THE EQUATIONS
Work on the x and y directions on separate sides of the page.

The only force acting on an object in gravity and gravity acts only in the y direction and it acts down. Thus:

a_y= -g (where g=9.8m/s)

Integrate to get
v_y = -gt + C (C a constant)

use intial conditions to find C, when t=0, v_y = vsinθ

THUS
v_y = -gt + vsinθ (THIS IS thus v = at + u OR a = (v-u)/t)

Integrate to get
y = -gt²/2 + vtsinθ + C (C another constant)

Use initial conditions to get (assuming here we are on the horizontal), when t=0, y=0

THUS
y = -gt²/2 + vtsinθ (THIS IS v = 1/2at² + ut and YOUR r = 0.5(2u + at)t equation - which is a weird way to write it))

[If we were on the cliff, then
y = -gt²/2 + vtsinθ + 100
Which is why this method is better for these sorts of weirder questions because you then don't need to think too hard about stuff when using the equations.]

NOW, to the x-direction. There is no acceleration in the x-direction. Thus:

a_x = 0

Integrate
v_x = C = vcosθ

Integrate again
x = vtcosθ + C
Use initial conditions, t=0, x=0

THUS
x = vtcosθ (THIS is s = ut)

Your equations are then:
y-direction
a_y= -g (where g=9.8m/s)
v_y = -gt + vsinθ
y = -gt²/2 + vtsinθ OR y = -gt²/2 + vtsinθ + 100 (depending on where you start in the vertical direction)

x-direction
a_x = 0
v_x = vcosθ
x = vtcosθ
SO
Max height when v_y = 0
Max distance when y=0

And so on.

Now, if you have acceleration that is not constant, then when you integrate it you will have a "t" term in there already. So if

a = 6t

Then v = 3t² + C

This is Maths Ext 2 stuff.

Now, if acceleration is constant, then you can do:

a = (v-u)/t

becasue (v-u) is just the linear change in velocity = delta(v)