Answer:
I believe s = 8t + 46 will work.
Step-by-step explanation:
We have2 data points plus the information that the car speed increases linearly with respect to time, so we should be able to find a linear equation with this information. We want an equation in the point-slope form: y=mx+b, where m is the slope and b is the y-intercept (the value of y when x = 0).
We can obtain the slope by calculating the Rise/Run between the two given data points:
Time is in seconds (sec) and speed in in miles per hour (mph).
Time (sec) Speed (mph)
3 70
8 110
The Rise/Run is the slope:
Rise (the change in Speed) is = (110-70) = 40mph
Run (the change in time) is = (8-3) = 5 seconds
Slope, or Rise/Run) = 40mph/5 sec or 8 mph/sec
We can write
y = 8x + b, where the slope, 8, is in mph/sec.
We don't have b, the speed at time = 0, and I initially thought "EASY!," the car isn't moving at the start. But I realized I was wrong, since if we are increasing speed by 8mph for every second, it isn't possible that the car was already at 70mph after only 3 seconds. And we are told it is a linear equation, so it precludes peeling out at very high speeds and then slowing down. I can only assume the car was already moving at time = 0 seconds; and the problems doesn't state the car was stationary when it started.
We can find the speed at t=0 seconds by using one of the data points in the equation and solving for b:
y = 8x + b (I'll use point (3,110))
110 = 8*(3) + b
b = 86
This means the car was travelling at 86mph when the data collection a=was initiated. What are the cops when you need them?
y = 8x + 86
To express this in the terms requested by the question:
s = 8t + 86mph
where s is speed in mph and t is time in seconds.
