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

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

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Answer 1 · 2018-04-20T05:56:06

The average speed is $= 2 m s^{- 1}$ $=2ms^-1$

Explanation:

The speed is the integral of the acceleration

$a (t) = 2 t^{2} - 2 t + 2$ $a(t)=2t^2-2t+2$

$v (t) = \int (2 t^{2} - 2 t + 2) d t$ $v(t)=int(2t^2-2t+2)dt$

$= \frac{2}{3} t^{3} - t^{2} + 2 t + C$ $=2/3t^3-t^2+2t+C$

Plugging in the initial conditions

$v (0) = 0$ $v(0)=0$

$v (0) = 0 - 0 + 0 + C$ $v(0)=0-0+0+C$

$C = 0$ $C=0$

The average speed on the interval $[0, 2]$ $[0,2]$ is

$(2 - 0) ¯ v = \int_{0}^{2} (\frac{2}{3} t^{3} - t^{2} + 2 t) d t$ $(2-0)barv=int_0^2(2/3t^3-t^2+2t)dt$

$2 ¯ v = {[\frac{1}{6} t^{4} - \frac{1}{3} t^{3} + t^{2}]}_{0}^{4}$ $2barv=[1/6t^4-1/3t^3+t^2]_0^2$

$= (\frac{8}{3} - \frac{8}{3} + 4) - (0)$ $=(8/3-8/3+4)-(0)$

$2 ¯ v = 4$ $2barv=4$

$¯ v = 2 m s^{- 1}$ $barv=2ms^-1$

Answer 2 · 2018-04-20T06:12:14

Given the acceleration of the object at t th instant as

$a (t) = 2 t^{2} - 2 t + 2$ $a(t) = 2t^2-2t+2$

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

$\Rightarrow v (t) = \int (2 t^{2} - 2 t + 2) d t + c$ $=>v(t)= int(2t^2-2t+2)dt+c$ , where c is integration constant.

$\Rightarrow v (t) = \frac{2}{3} t^{3} - \frac{2}{2} t^{2} + 2 t + c$ $=>v(t)= 2/3t^3-2/2t^2+2t+c$

Given that object is still at $t = 0$ $t=0$ , we get $v (0) = 0$ $v(0)=0$

Hence $v (0) = 0 = \frac{2}{3} \times 0 - \frac{2}{2} \times 0 + 2 \times 0 + c$ $v(0)= 0=2/3xx0-2/2xx0+2xx0+c$

So $c = 0$ $c=0$
Hence
$v (t) = \frac{2}{3} t^{3} - t^{2} + 2 t$ $v(t)= 2/3t^3-t^2+2t$

$\Rightarrow \frac{d s (t)}{d t} = \frac{2}{3} t^{3} - t^{2} + 2 t$ $=>(ds(t))/(dt)= 2/3t^3-t^2+2t$

So displacement in the time interval $t \in [0, 2]$ $tin[0,2]$
$\Rightarrow s (t) = \int_{0}^{2} (\frac{2}{3} t^{3} - t^{2} + 2 t) d t$ $=>s(t)=int_0^2( 2/3t^3-t^2+2t)dt$

$\Rightarrow s (t) = {[\frac{2}{12} t^{4} - \frac{1}{3} t^{3} + \frac{2}{2} t^{2}]}_{0}^{4}$ $=>s(t)=[ 2/12t^4-1/3t^3+2/2t^2]_0^2$

$\Rightarrow s (t) = [\frac{2}{12} \cdot 2^{4} - \frac{1}{3} \cdot 2^{3} + 2^{2}]$ $=>s(t)=[ 2/12*2^4-1/3*2^3+2^2]$

$\Rightarrow s (t) = [\frac{8}{3} - \frac{8}{3} + 4] = 4$ $=>s(t)=[ 8/3-8/3+4]=4$ unit

Hence average velocity $¯ v = \frac{displacement}{time interval} = \frac{4}{2} = 2$ $barv="displacement"/"time interval"=4/2=2$ unit/s

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

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