MHB Velocity Time graphs: Find acceleration in the first 15 mins in km/h^2

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To find acceleration from a velocity-time graph, the formula used is a = (v(9.25) - v(9)) / (0.25 - 0), indicating uniform acceleration from 9:00 to 9:15. The distance traveled between 9 and 11 can be calculated using the integral of velocity, d = ∫9^11 v(t) dt, which represents the area under the curve. Different geometric shapes are considered to calculate this area, including triangles and rectangles, with specific dimensions leading to various distance values. The average velocity over the interval is calculated using the formula |v̄| = (1 / (11-9)) ∫9^11 v(t) dt. The discussion emphasizes understanding the relationship between velocity, area under the curve, and distance traveled.
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(a) note acceleration is uniform from 9:00 to 9:15 ...

$a = \dfrac{v(9.25) - v(9)}{.25 - 0}$

(b) since $v \ge 0$ for $9 \le t \le 11$, $\displaystyle d = \int_9^{11} v(t) \, dt$

(c) $\displaystyle |\bar{v}| = \dfrac{1}{11-9} \int_9^{11} v(t) \, dt$
 
For (b) and (c) that integral is the "area under the curve". For b, I would think of this as:
1) a triangle with base "1/4 hour" and height "50 km/hr". The area of that triangle is (1/2)(1/4)(50)= 6.25 km.
2) a rectangle with base "1/2 hour" and height "50 km/hr". The area of that rectangle is (1/2)(50)= 25 km.
3) a trapezoid with bases "50 km/hr" and "100 km/hr" and height "1/4 hour" (yes, I've swapped "height" and "base" to better fit the trapezoid). The area is (1/2)(50+ 100)(1/4)= 18.75 km.
4) a rectangle with base "1/2 hr" and height "100 km/hr". The area is (1/2)(100)= 50 km.
5) a triangle with base "1/2 hr" and height "100 km". The area is (1/2)(1/2)(100)= 25 km.
 
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