Question

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)

Answer 1

To 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 2

x - (1/3)y = -44/3

Explanation:

Let's call:

x: time

y: velocity of the car

The point-slope form is:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is a point in the line

The standard form is:

ax + by = c

(where a > 0)

We know two points of the line, (3, 53) and (6, 62). The slope is calculated as follows:

m = (62 -53) / (6-3) = 3

Replacing into the equation:

y - 53 = 3(x - 3)

y - 53 = 3x - 9

-3x + y = -9 + 53

(-3x + y)/(-3) = (-9 + 53)/(-3)

x - (1/3)y = -44/3