What is the average speed of an object that is not moving at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 6 t - 9$ $a(t) =6t-9$ on $t \in [3, 5]$ $t in [3, 5]$ ?

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Answer 1 · 2016-01-03T00:59:15

Take the differential definition of acceleration, derive a formula connecting speed and time, find the two speeds and estimate the average.

$u_{a v} = 15$ $u_(av)=15$

The definition of acceleration:

$a = \frac{d u}{d t}$ $a=(du)/dt$

$a \cdot d t = d u$ $a*dt=du$

$\int_{0}^{t} a (t) d t = \int_{0}^{u} d u$ $int_0^ta(t)dt=int_0^udu$

$\int_{0}^{t} (6 t - 9) d t = \int_{0}^{u} d u$ $int_0^t(6t-9)dt=int_0^udu$

$\int_{0}^{t} (6 t \cdot d t) - \int_{0}^{t} 9 d t = \int_{0}^{u} d u$ $int_0^t(6t*dt)-int_0^t9dt=int_0^udu$

$6 \int_{0}^{t} (t \cdot d t) - 9 \int_{0}^{t} d t = \int_{0}^{u} d u$ $6int_0^t(t*dt)-9int_0^tdt=int_0^udu$

$6 \cdot {[\frac{t^{2}}{2}]}_{0}^{t} - 9 \cdot {[t]}_{0}^{t} = {[u]}_{0}^{u}$ $6*[t^2/2]_0^t-9*[t]_0^t=[u]_0^u$

$6 \cdot (\frac{t^{2}}{2} - \frac{0^{2}}{2}) - 9 \cdot (t - 0) = (u - 0)$ $6*(t^2/2-0^2/2)-9*(t-0)=(u-0)$

$3 t^{2} - 9 t = u$ $3t^2-9t=u$

$u (t) = 3 t^{2} - 9 t$ $u(t)=3t^2-9t$

So the speed at $t = 3$ $t=3$ and $t = 5$ $t=5$ :

$u (3) = 3 \cdot 3^{2} - 9 \cdot 3 = 0$ $u(3)=3*3^2-9*3=0$

$u (5) = 30$ $u(5)=30$

The average speed for $t \in [3, 5]$ $t in[3,5]$ :

$u_{a v} = \frac{u (3) + u (5)}{2}$ $u_(av)=(u(3)+u(5))/2$

$u_{a v} = \frac{0 + 30}{2}$ $u_(av)=(0+30)/2$

$u_{a v} = 15$ $u_(av)=15$

What is the average speed of an object that is not moving at t=0t=0 and accelerates at a rate of a(t)=6t−9a(t) =6t-9 on t∈[3,5]t in [3, 5]?