Question

HELP PLEASE PLEASE PLEASE

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 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 seven hours?

Answer 1

Please I really need help with this

Answer 2

that means (2, 54) and (4, 58) are two points on the required graph and u can use these to write the equation for velocity in slope-intercept form

Answer 3

start by finding the slope

Answer 4

how do I do that

Answer 5

please u really need help with this

Answer 6

I*

Answer 7

Hi!!

Answer 8

hi

Answer 9

slope is \[\frac{y_2-y_1}{x_2-z_1}=\frac{62-53}{6-3}\]

Answer 10

i get \(3\)

Answer 11

ok so i put all of this down?

Answer 12

yes, then you still have to find the equation \[y-y_1=m(x-x_1)\] or \[y-53=3(x-3)\]

Answer 13

ok

Answer 14

multiply out, get \[y-53=3x-9\]add \(53\) and your final answer is \[y=3x+44\]

Answer 15

o answer the question above, determine the velocity of the car at time zero. To do such, determine the increase in velocity every hour (acceleration).

a = (62 - 53) / (6 -3) = 3 km/ h^2

Calculate for the initial velocity (v) by using either of the velocities paired with the corresponding time,

v = 53 km/h - (3 km/h^2) x 3 h = 44 km/h

The equation is y = 44 + 3x. Where x is the time in hours, y is the velocity in km/h.

Answer 16

Thankyou so much can you help me with Part B?

Answer 17

sure
you got \[y=3x+44\]put \(x=7\) to get \[y=3\times 7+44\] whatever that is

Answer 18

thnx can you help me with one more?