A car acclerates at 4m/s/s for 5s and then drove at its final velocity for 8s. It then decelerates at 2m/s/s for 2s. Draw a velocity-time graph and use the graph to find the distance covered during the 15s described.
Would I plot (5,8) and then draw a line from the orgin to that point to draw the graph?
I know that the slope of a velocity-time grpah is acceleration but I don't know how I would find the distace using the graph in the 15s described.
If it's accelerating the velocity is a line with positive slope. If the velocity is constant the line is horizontal.
You wrote
"I know that the slope of a velocity-time grpah is acceleration"
For x=0 to x=5 it has slope 4. From x = 5 to x=13 slope is 0. Deceleration has a negative slope.
You asked
"Would I plot (5,8) and then draw a line from the orgin to that point to draw the graph?"
No. Determine the point (0,0) and draw a line with slope 4 through it to (5,20). Then the horizontal line, then the one with negative slope.
Time is on the x-axis, velocity on the y-axis.
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