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poizon [28]
3 years ago
5

Which equations can be used to solve for acceleration? Check all that apply.

Mathematics
2 answers:
Masteriza [31]3 years ago
5 0

Equations can be used to solve for acceleration

\rm t=\dfrac{\Delta v }{a}

\rm vi=vf-at

\rm vf=at+vi

\rm a=\dfrac{\Delta v }{t}

<h3>Further explanation </h3>

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

• An equation of uniformly accelerated motion

\large {\boxed {\bold {x = xo + vo.t + \dfrac {1} {2} at ^ 2}}}

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

\displaystyle a = \dfrac {v2-v1} {t2-t1}

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

<h3>Learn more </h3>

The distance of the elevator

brainly.com/question/8729508

resultant velocity

brainly.com/question/4945130

the average velocity

brainly.com/question/5248528

salantis [7]3 years ago
3 0
a =\frac{v_f - v_i}{t}\\\\at = v_f -v_i\\\\\boxed{\bf{v_f =at +v_i}}\\\\\boxed{\bf{v_i = v_f-at}}
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Does 2 1/3-1 3/4 need require renaming
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Answer:

\frac{7}{12}

Step-by-step explanation:

2 \frac{1}{3}  - 1 \frac{3}{4}

The fractions above are known as mixed numbers (that is, fractions that contain the combination of a whole number and proper fraction). The mixed numbers can equally be renamed as improper fractions. However, renaming can only occur the the mixed numbers are being transformed as below:

\frac{7}{3}  -  \frac{7}{4}

The transformation occur through a simple approach thus: for the first mixed number; 2 1/3 = 2 multiplied by 3 plus 1 divide by 3 equals 7/3. Similarly, mixed number 1 3/4 is transformed by this: 1 multiplied by 4 plus 3 divide by 4 equals 7/4.

Simplifying further to arrive at final result implies,

\frac{7}{3}  -  \frac{7}{4}

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3 years ago
Slope Intercept Form And Equations From Graphs
r-ruslan [8.4K]

Answer:

1. Choice B

2.  Choice C

3.  X=-2

4.  Choice A

5. y = 5/2 x + 5

Step-by-step explanation:

1.  

   the slope is down 3 over 2     m =-3/2

point slope form

y-y1= m (x-x1)

y-y1 = -3/2 (x-x1)

Choice B


2.  (-3,-1)  (1/2,2)

slope = (y2-y1)/(x2-x1)

m=(2--1)/(1/2--3)  = (2+1)/(1/2+3) = 3/(3.5) =  multiply top and bottom by 2

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point slope form

y-y1 = m(x-x1)

y--1 = 6/7 (x--3)

y+1 =6/7 (x+3)

multiply by 7

7y+7 = 6(x+3)

7y+7 = 6x+18

subtract 7y from each side

7 = 6x-7y+18

subtract 18 from each side

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Choice C


3.  x=-2

This is a vertical line, the value of x never changes


4.   (-1,2) (1,-4)

slope =(y2-y1)/(x2-x1)

      = (-4-2)/(1--1) = -6/(1+1) = -6/2 = -3

point slope form

y-y1 = m(x-x1)

y-2 = -3(x--1)

y-2 = -3(x+1)

y-2 = -3x-3

add 2 to each side

y = -3x-1

Choice A


5.  (-2,0)  (0,5)

the y intercept is 5

slope is change in y over change in x

slope =(y2-y1)/(x2-x1)  = (0-5)/(-2-0) = -5/-2 = 5/2

slope intercept form

y= mx+b

y = 5/2 x + 5



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