Which equations can be used to solve for acceleration? check all that apply.
t =
vf = at – vi
a =
vi = vf – at
v =
vf = at + vi
a =

Equations can be used to solve for acceleration

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration

• An equation of uniformly accelerated motion

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

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

The distance of the elevator

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resultant velocity

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the average velocity

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the circle graph is not a function.

step-by-step explanation:

the graph of the circle is not a function.

because for a graph to be a function, each value of x must give one value of y

in the circle, the values of x for -2< x< 2 each give 2 values of y.

hence this is not a function.