What is the average speed of an object that is still at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = \frac{t}{6}$ $a(t) =t/6$ from $t \in [0, 1]$ $t in [0, 1]$ ?

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Answer 1 · 2016-01-04T17:10:23

You also need the initial speed of the object $u_{0}$ $u_0$ . The the answer is:

$u_{a v} = 0.042 + u_{0}$ $u_(av)=0.042+u_0$

Definition of acceleration:

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

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

$\int_{0}^{t} a (t) d t = \int_{u_{0}}^{u} d u$ $int_0^ta(t)dt=int_(u_0)^udu$

$\int_{0}^{t} (\frac{t}{6}) d t = \int_{u_{0}}^{u} d u$ $int_0^t(t/6)dt=int_(u_0)^udu$

$\frac{1}{6} \int_{0}^{t} (t) d t = \int_{u_{0}}^{u} d u$ $1/6int_0^t(t)dt=int_(u_0)^udu$

$\frac{1}{6} (\frac{t^{2}}{2} - \frac{0^{2}}{2}) = u - u_{0}$ $1/6(t^2/2-0^2/2)=u-u_0$

$u (t) = \frac{t^{2}}{12} + u_{0}$ $u(t)=t^2/12+u_0$

To find the average speed:

$u (0) = \frac{0^{2}}{12} + u_{0} = u_{0}$ $u(0)=0^2/12+u_0=u_0$

$u (1) = \frac{1^{2}}{12} + u_{0} = \frac{1}{12} - u_{0}$ $u(1)=1^2/12+u_0=1/12-u_0$

$u_{a v} = \frac{u_{0} + u_{1}}{2}$ $u_(av)=(u_0+u_1)/2$

$u_{a v} = \frac{u_{0} + \frac{1}{12} + u_{0}}{2}$ $u_(av)=(u_0+1/12+u_0)/2$

$u_{a v} = \frac{2 u_{\oplus} \frac{1}{12}}{2}$ $u_(av)=(2u_o+1/12)/2$

$u_{a v} = \frac{2 u_{0}}{2} + \frac{\frac{1}{12}}{2}$ $u_(av)=(2u_0)/2+(1/12)/2$

$u_{a v} = u_{0} + \frac{1}{24}$ $u_(av)=u_0+1/24$

$u_{a v} = 0.042 + u_{0}$ $u_(av)=0.042+u_0$

What is the average speed of an object that is still at t=0t=0 and accelerates at a rate of a(t)=t6a(t) =t/6 from t∈[0,1]t in [0, 1]?