Please help or I might get in big trouble for getting a terrible grade

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.

answer: procent 54 wynosi 33,33%