Question

Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 54 km/h. After four hours, the velocity of the car is 58 km/h.

Part A: Write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used. (5 points)

Part B: How can you graph the equation obtained in Part A for the first six hours? (5 points)

Answer 1

Do u get it?

Answer 2

Does it say anywhere to assume linear relationship between the velocity of the car and the time? Is this question in the chapter that deals with equations of a straight line?

Answer 3

Well it depends on what the problem is giving you sometimes it can be a straight line.

Answer 4

@aum

Answer 5

@nincompoop Hi could you please help me?

Answer 6

@FreeWilly Hi

Answer 7

hello???

Answer 8

@amistre64 Hi Sorry but could you please help me with this?

Answer 9

@Ashleyisakitty could u please help me?

Answer 10

the broken nature of the site is making tags and notifs hard to get thru

Answer 11

how do we write an equation of a line when we are just given 2 points of reference?

Answer 12

Thank You finally someone who answered me!

Answer 13

I have no idea im sorry i dont get how to solve velocity

Answer 14

its not really about solving velocity, but modeling with the data given

Answer 15

we are given 2 points of reference, the simplest model would be a straight line between the 2 points: a linear model

Answer 16

ok

Answer 17

to do that we need one of the points, and the slope between the points in order to develop a line equation

Answer 18

we could work a system of equations as well but that may be more advanced then your course calls for

Answer 19

in the problem it said after 2 hours the cars was 54 km/h and after 4 hours it was 58 km/h so to me it increased by 4 every 2 hours

Answer 20

thats a good observation, that rate of change is called a slope in a linear equation

Answer 21

yep

Answer 22

velocity is equal to, (observed rate of change) times (hrs - a), plus b

v = m(t-a) + b

but then this isnt in standard form

standard form of a line is: Ax + By = C, but we can always rework it into this

Answer 23

so v = 4/2 (t - 2) + 54 is called point slope format

Answer 24

4/2 = 2/1 sooo

v = 2(t-2)+54

v = 2t -4 +54

-2t +v = 50

but standard form start with a postive so adjust this by a factor of -1

2t -v = -50 would seem fair to me

Answer 25

graphing it has many approaches as well, the intercept method might be suitable

Answer 26

ok

Answer 27

find the point t=0 v=? and plot it

find the point t=6 v=? and plot it

then draw the line between the 2 points

Answer 28

ok i get it now Thanks for the GREAT help!