The integral represents the area between the curve and the x-axis.

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Presentation transcript:

The integral represents the area between the curve and the x-axis. Part (a) At t=0, the particle is at x=-2. 8 3 2 Keep in mind that the velocity is graphed, so position would be the integral. The integral represents the area between the curve and the x-axis.

The particle is farthest to the left at this point. Part (a) At t=0, the particle is at x=-2. -2 – 8 = -10 8 3 2 At t=3, the particle is at x=-10. -10 + 3 - 2 = -9 At t=6, the particle is at x=-9. The two possible locations would be at t=3 and t=6, because you’re following an interval where v(t)<0. The particle is farthest to the left at this point.

At t=0, the particle is at x=-2. Part (b) At t=0, the particle is at x=-2. 8 3 2 The motion of the particle would look like this... -8 -6 -4 -2 2 4 6 8 -10

At t=0, the particle is at x=-2. At t=3, the particle is at x=-10. Part (b) At t=0, the particle is at x=-2. 8 3 2 The motion of the particle would look like this... At t=3, the particle is at x=-10. -8 -6 -4 -2 2 4 6 8 -10

The motion of the particle would look like this... Part (b) At t=0, the particle is at x=-2. 8 3 2 The motion of the particle would look like this... At t=3, the particle is at x=-10. At t=5, the particle is at x=-7. -8 -6 -4 -2 2 4 6 8 -10

As you can see, the particle was at x = -8 three times. Part (b) At t=0, the particle is at x=-2. 8 3 2 The motion of the particle would look like this... At t=3, the particle is at x=-10. At t=5, the particle is at x=-7. At t=6, the particle is at x=-9. As you can see, the particle was at x = -8 three times. -8 -6 -4 -2 2 4 6 8 -10

Part (c) The velocity is negative during this interval, but the acceleration (the slope, in this case) is positive. Therefore, the SPEED is DECREASING. In addition, the graph of v(t) is approaching the x-axis, which also indicates the particle is slowing down.

Part (d) Since acceleration is the slope of the v(t) graph, a < 0 on the intervals (0,1) and (4,6).