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**HallsofIvy** In more advanced mathematics (Calculus) we would say that the total distance traveled is the **integral** of the velocity function. That is also the "area under the curve" (you say you know that) and here, that's easy- it is the area of a trapezoid with bases of length 100 and 40 and height 8: $\displaystyle \frac{100+ 40}{2}(8)= 70(8)= 560$ m. Perhaps your tm thought you had learned the area for a trapezoid long ago. Or you can do what skeeter suggested in his last post: the trapezoid can be seen as two triangles and a rectangle. Find the area of each and add.

The bus has three different accelerations, one of which is 0. Are you asked for all three or the **average** acceleration. If "average" this is trivial: it starts with 0 velocity and ends with 0 velocity!

Finally, for the last part, you are told that T< 40 so you are only looking at the first part of the graph. The bicycle travels at a constant 5 mph (another of those damn bicyclists who doesn't stop for red lights!) so after time T has gone 5t miles. That's exactly the same as saying that if you drop a perpendicular at time T, you have a rectangle of height 5 and base T and so area 5T. The bus, during that time gets up to a speed of 8 m/s in 40 s and so has an acceleration of 40/8= 5m/s^2. The distance gone by the bus is a triangle with base T and height 5T. Find the area of that triangle, set it equal to 5T and solve for T.