What is the average speed of an object that is still at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 12 - 2 t$ $a(t) =12-2t$ from $t \in [3, 6]$ $t in [3, 6]$ ?

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Answer 1 · 2016-11-08T02:56:42

Given
Acceleration $a (t) = 12 - 2 t$ $a(t)=12-2t$

$\Rightarrow \frac{d (v (t))}{d t} = 12 - 2 t$ $=>(d(v(t)))/(dt)=12-2t$

$\Rightarrow \int (d (v (t))) = \int (12 - 2 t) d t$ $=>int(d(v(t)))=int(12-2t)dt$

$\Rightarrow v (t) = 12 t - 2 \times \frac{t^{2}}{2} + c$ $=>v(t)=12t-2xxt^2/2+c$

$\Rightarrow v (t) = 12 t - t^{2} + c$ $=>v(t)=12t-t^2+c$

where t = integration constant

Now by the given condition

at t=0, v(0)=0

So $v (0) = 0 = 12 \times 0 - 0^{2} + c \Rightarrow c = 0$ $v(0)=0=12xx0-0^2+c=>c=0$

Hence

$\Rightarrow v (t) = 12 t - t^{2}$ $=>v(t)=12t-t^2$

$\Rightarrow \frac{d (s (t))}{d t} = 12 t - t^{2}$ $=>(d(s(t)))/(dt)=12t-t^2$

$\Rightarrow \int_{3}^{6} (d (s (t))) = \int_{3}^{6} (12 t - t^{2})$ $=>int_3^6(d(s(t)))=int_3^6(12t-t^2)$

$\Rightarrow s (6) - s (3) = {[6 t^{2} - \frac{t^{3}}{3}]}_{3}^{2}$ $=>s(6)-s(3)=[6t^2-t^3/3]_3^6$

$= 6 \cdot 6^{2} - \frac{6^{3}}{3} - 6 \cdot 3^{2} + \frac{3^{3}}{3}$ $=6*6^2-6^3/3-6*3^2+3^3/3$

$= 216 - 72 - 54 + 9 = 99$ $=216-72-54+9=99$

Average speed in time [3,6] is

$= \frac{s (6) - s (3)}{6 - 3} = \frac{99}{3} = 33 u n i t$ $=(s(6)-s(3))/(6-3)=99/3=33 unit$

What is the average speed of an object that is still at t=0t=0t=0 and accelerates at a rate of a(t) =12-2ta(t)=12−2ta(t) =12-2t from t in [3, 6]t∈[3,6]t in [3, 6]?