need help with both questions!

[SOLVED] need help with both questions!
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#14 isn't really a Physics problem.  It's more of just reading a graph.A). When speed changes, acceleration is       (change in speed) / (time for the change) .To be correct about it, acceleration can be positive ... when speed is increasing ... or it can be negative ... when speed is decreasing. So, on this graph, there are two periods of acceleration:From zero to 2 seconds, acceleration = (8 m/s) / (4 sec) = 2 m/s² .From 10 to 12 seconds, acceleration = (-4 m/s) / (2 sec) = -2 m/s² .B). From 12 to16 seconds, you can read the speed right from the graph.  It's 4 m/s .C).  From 2 to 10 seconds, the objects speed is a steady 8 m/s.Covering 8 m/s every second for 8 seconds, it covers 64 meters.Do you remember that distance is the area under the speed/time graph?  You can see that plainly on this graph.  From 2 to 10 sec,there are 16 blocks.  Each block is (2 m/s) high and (2 sec) wide,so its area is (2 m/s) x (2 sec) = 4 meters.  The area of 16 blocksis (16) x (4 meters) = 64 meters.====================================#15.a).  constant velocity on a distance graph is a line that slopes up;constant velocity on a velocity graph is a horizontal line;b). positive constant acceleration on a distance graph is aline that curves up;positive constant acceleration on a velocity graph is astraight line that slopes up;c).  "uniformly slowing down to a stop" on a distance graph is a line that's less and less curved as time goes on, and eventually reaches the x-axis. "uniformly slowing down to a stop" on a velocity graph is a straight line that slopes down, and stops when it reaches the x-axis.

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