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

[hadiahabibi]

Helpp meee pleaseee View attachment 8974

 

[skeeter]

(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) $\displaystyle |\bar{v}| = \dfrac{1}{11-9} \int_9^{11} v(t) \, dt$

 

[HallsofIvy]

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.