What is the average speed of an object that is moving at $2 \frac{m}{s}$ $2 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 - t$ $a(t) =5-t$ on $t \in [0, 3]$ $t in [0,3]$ ?

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What is the average speed of an object that is moving at $2 \frac{m}{s}$ $2 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 - t$ $a(t) =5-t$ on $t \in [0, 3]$ $t in [0,3]$ ?

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Answer 1 · 2016-04-20T13:29:43

Given acceleration $a (t) = 5 - t$ $a(t)=5-t$
OR, $\frac{d v}{d t} = 5 - t$ $(dv)/(dt)=5-t$
, $\Rightarrow d v = 5 d t - t d t$ $=>dv=5dt-tdt$
, $\Rightarrow \int d v = 5 \int d t - \int t d t + c$ $=>intdv=5intdt-inttdt+c$ ,where c= integration constant
$\Rightarrow v = 5 t - \frac{t^{2}}{2} + c$ $=>v=5t-t^2/2+c$ ....(1)
Given at $t = 0, v = 2 \frac{m}{s}$ $t=0 ,v=2m/s$
So $c = 2$ $c = 2$
and eq(1) becomes
$\Rightarrow v = 5 t - \frac{t^{2}}{2} + 2$ $=>v=5t-t^2/2+2$

Now distance traversed during $t = 0 \to t = 3 s$ $t=0 to t= 3s$

$s = \int_{0}^{3} v d t = \int_{0}^{3} (5 t - \frac{t^{2}}{2} + 2) d t$ $s=int_0^3vdt=int_0^3(5t-t^2/2+2)dt$
$= {[5 \frac{t^{2}}{2} - \frac{t^{3}}{6} + 2 t]}_{0}^{3} = 5 \cdot \frac{3^{2}}{2} - \frac{3^{3}}{6} + 2 \cdot 3 = 24 m$ $=[5t^2/2-t^3/6+2t]_0^3=5*3^2/2-3^3/6+2*3=24m$
Hence Average speed
$= \frac{s}{t} = \frac{24}{3} = 8 \frac{m}{s}$ $=s/t=24/3=8m/s$

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What is the average speed of an object that is moving at $2 \frac{m}{s}$ $2 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 - t$ $a(t) =5-t$ on $t \in [0, 3]$ $t in [0,3]$ ?

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What is the average speed of an object that is moving at $2 \frac{m}{s}$ $2 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 - t$ $a(t) =5-t$ on $t \in [0, 3]$ $t in [0,3]$ ?