Question

medal and fan!!!

Answer 1

whats the question

Answer 2

The table below represents the velocity of a car as a function of time:

Answer 3

Time
(hour)
x Velocity
(miles/hours)
y
0 50
1 52
2 54
3 56

Answer 4

late for class i'll try to come back soon

Answer 5

Part A: What is the y-intercept of the function, and what does this tell you about the car? (4 points)

Part B: Calculate the average rate of change of the function represented by the table between x = 1 to x = 3 hours, and tell what the average rate represents. (4 points)

Part C: What would be the domain of the function if the velocity of the car was measured until it reached 60 miles/hour and the car does not change motion? (2 points)

Answer 6

@Keigh2015 @Michele_Laino @tkhunny can you help please?

Answer 7

here we have to write the equation of the function between speed and time

Answer 8

okay how do we do that?

Answer 9

we have to strat from the general equation:
\(v=at+b\)
where \(v\) is the speed, and \(t\) is time

Answer 10

i dont understand

Answer 11

hint:
let's consider the first point \((0,50)\), which means that when t=0, then v=50, so I can rewrite my equation like below:
\(50=a \cdot 0+b\)
what is \(b\) ?

Answer 12

b=50

Answer 13

correct! so we can update my equation as follows:
\(v=at+50\)

Answer 14

next I consider the second point \((1,52)\) and, as before, I rewrite my equation with such point
\(52= a \cdot 1 +50\)
what is \(a\) ?

Answer 15

a=2

Answer 16

that's right!! :)
so the equation is:
\(v=2t+50\)
now, what are the slope, and the y-intercept ?

Answer 17

hint:

Answer 18

sorry i went to get something to eat

Answer 19

oh okay i get it i think

Answer 20

for part B, we have to compute this quantity:
\[\Large \frac{{v\left( 3 \right) - v\left( 1 \right)}}{{3 - 1}} = \frac{{56 - 52}}{2} = ...?\]

Answer 21

i have not been able to answer this

Answer 22

im really frustrated

Answer 23

it is simple:
\[\Large
\frac{{v\left( 3 \right) - v\left( 1 \right)}}{{3 - 1}} = \frac{{56 - 52}}{2} = \frac{4}{2} = 2\]
which represents an average acceleration
Now, I have to go out with my car, I'm sorry!

Answer 24

wich part did we do

Answer 25

the third part, namely part C

Answer 26

okay so did we finish the question?

Answer 27

no, I'm sorry I have to go out, I will try to return as soon as possible

Answer 28

k thank you

Answer 29

:)

Answer 30

Ok, Well I cant solve this problem, But @ganeshie8 @Luigi0210 @mayankdevnani @dumbcow @dan815 @Awolflover1 can. Good luck bro.

Answer 31

@zepdrix can you help me with the last part of this please?

Answer 32

@TrojanPoem can you help please?

Answer 33

i just need help with part c

Answer 34

The function :

v = v0 + at



v0 = 50 m/h ( from the table)

at t = 1 , v = 52

52 = 50 + a

a = 2

so the function is:

v= v0 + at

v = 50 + 2t

the y-intercept is the initial velocity (at the beginning of studying the motion)


the maximum value for the domain is 60

now we

0.5 v - 25 = t

t > 0


0.5v > 25
v > 50


so the range is [50,60]

now to find the domain


find t when v = maximum = 60

60 = 50 + 2t

10 = 2t

t = 5 second

domain [0,5]

Answer 35

is this part C ?

Answer 36

I solved it all.

Answer 37

what do you mean?

Answer 38

(C) domain [0,5]

Answer 39

can you tell me what part c is?

Answer 40

i have part A/B

Answer 41

find t when v = maximum = 60

60 = 50 + 2t

10 = 2t

t = 5 second

domain [0,5]

Answer 42

that is all there is for part c?

Answer 43

Yeah

Answer 44

oh okay then! thank you!!

Answer 45

Any time.

Answer 46

If you have solved this question you need to close it, ok. I dont mean to be rude just saying.