Equations can be used to solve for acceleration
[tex]\rm t=\dfrac{\Delta v }{a}[/tex]
[tex]\rm vi=vf-at[/tex]
[tex]\rm vf=at+vi[/tex]
[tex]\rm a=\dfrac{\Delta v }{t}[/tex]
Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration
• An equation of uniformly accelerated motion
[tex] \large {\boxed {\bold {x = xo + vo.t + \dfrac {1} {2} at ^ 2}}} [/tex]
V = vo + at or vf = vi + at
Vt² = vo² + 2a (x-xo)
x = distance on t
vo / vi = initial speed
vt / vf = speed on t / final speed
a = acceleration
Acceleration is a change in speed within a certain time interval
a = Δv / Δ t
[tex] \displaystyle a = \dfrac {v2-v1} {t2-t1} [/tex]
The velocity graph with respect to time (v-t) in this motion will be in the form of sloping straight lines with a gradient of tan θ = a
For acceleration that is positive (a> 0) is called accelerated motion, the graph is sloping upward, while for acceleration that is negative (a <0) is called slowed down motion, the graph is sloping downward
We complete the available answer choices
t = delta v over a.
vf = at - vi
a = d over t.
vi = vf - at
v = a over t.
vf = at + vi
a = delta v over t.
From the general equation of uniformly accelerated motion above, what is appropriate is:
1. t = delta v over a ---> can be used, a = v / t
2. vf = at - vi ---> cannot be used, it should be vf = vi + at
3. a = d over t .---> d = displacement has the formula: d = v x t, so this 3rd equation cannot be used
4. vi = vf - at ----> can be used
5. v = a over t ----> cannot be used, it should be v = a x t
6. vf = at + vi ---> can be used
7. a = delta v over t ----> can be used
So the equation that can be used to find acceleration is
1,4,6 and 7
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