Part (a) ½(4)(20) + (12)(20) ½(8)(20) +

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Presentation transcript:

Part (a) ½(4)(20) + (12)(20) ½(8)(20) + This represents the total distance (in meters) traveled by the car during the first 24 seconds. The integral is equal to the area under the graph. In this case, we’re adding two triangles and a rectangle. 24 v(t) dt = 24 v(t) dt = 40 + 240 + 80 360 meters

Part (b) m=0 m=-5/2 m=5 v’(4) does not exist. The reason why is that the limit as t approaches 4 from the left is 5, while the limit as t approaches 4 from the right is zero. Since the left- and right-hand limits are different, no overall limit exists at t=4. Therefore, there is no derivative there either. v’(20) equals -5/2 meters per second. We get this answer from the slope of the v(t) graph as it passes through t=20.

Acceleration is the derivative (slope) of the velocity graph. Part (c) Acceleration is the derivative (slope) of the velocity graph. a(t) = 5 when 0 < t < 4 0 when 4 < t < 16 -5/2 when 16 < t < 24 DNE when t=4 or t=16

Part (d) (8,20) (20,10) Since this function is piecewise, the only way to find the average rate of change is to use the slope formula from Algebra 1. Avg. rate of change = 10-20 20-8 = -5 6 m/sec Normally, the Mean Value Theorem would tell us that the slope of the curve is -5/6 somewhere between t=8 and t=20. From Part (c), however, we know that the slope is never actually equal to -5/6 anywhere.

Part (d) (8,20) (20,10) The reason that the Mean Value Theorem can’t guarantee us a slope of -5/6 is because the MVT can only be used on those portions of functions that are fully differentiable. Remember, our function has no derivative at t=16. Therefore, the MVT can’t be used.