How do you set up this?I am lost

Find an equation for the linear model of the situation below and use it to make a prediction. A train is traveling north at a constant rate. At 3:00 P.M. it is 55 miles north of a city. At 4:15 P.M. it is 80 miles north of the city. If d represents the distance in miles, and t represents the time in hours, how many miles north of the city will the train be at 5:45 P.M.?

x = xi + v t

80 = 55 + v(75 minutes)
v = (80-55)/75 = (1/3) mile/min
so
x = 55 + (1/3) t if t in minutes
or
x = 55 + (20) t if t in hours

at 5:45 t = 2:45 = 2.75 hours

x = 55 + 20(2.75) = 110 miles

My answer book had

64t+ 55
d= 231
I don't understand how they came up with these #'s

To find an equation for the linear model, you need to first identify the variables involved and determine how they are related. In this case, the variables are distance (d) and time (t).

We know that the train is traveling at a constant rate, which means the relationship between distance and time is linear. This implies that there is a constant rate of change, which we can define as the slope (m) of the linear model.

To find the slope, we can use the formula:
m = (change in y) / (change in x)

In this case, the change in distance is 80 miles - 55 miles = 25 miles, and the change in time is 4.25 hours - 3 hours = 1.25 hours. Therefore, the slope is:
m = 25 miles / 1.25 hours = 20 miles/hour

Now that we have the slope, we can use the point-slope form of a linear equation to find the equation of the line. The point-slope form is:
y - y1 = m(x - x1)

Let's use the first data point (3:00 P.M., 55 miles) as our reference point (x1, y1). Plugging in the values, we get:
d - 55 = 20(t - 3)

Now, we want to find the distance at 5:45 P.M., which is 2 hours and 45 minutes after 3:00 P.M. Convert 45 minutes to hours by dividing by 60:
2 hours + 45 minutes / 60 minutes per hour = 2.75 hours

Substitute t = 2.75 into the equation and solve for d:
d - 55 = 20(2.75 - 3)
d - 55 = -5
d = 55 - 5
d = 50

Therefore, at 5:45 P.M., the train will be 50 miles north of the city.