What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $8 \frac{m}{s}$ $8 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 t^{2} - 3 t$ $a(t) =5t^2-3t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $8 \frac{m}{s}$ $8 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 t^{2} - 3 t$ $a(t) =5t^2-3t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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Answer 1 · 2017-11-29T07:45:06

The average speed is $= 38.67 m s^{- 1}$ $=38.67ms^-1$

Explanation:

The speed is the integral of the acceleration

$a (t) = 5 t^{2} - 3 t$ $a(t)=5t^2-3t$

Then,

$v (t) = \int (5 t^{2} - 3 t) d t = \frac{5}{3} t^{3} - \frac{3}{2} t^{2} + C$ $v(t)=int(5t^2-3t)dt=5/3t^3-3/2t^2+C$

Plugging in the initial conditions at $t = 0$ $t=0$

$v (0) = 8 = 0 - 0 + C$ $v(0)=8=0-0+C$

$C = 8$ $C=8$

Therefore,

$v (t) = \frac{5}{3} t^{3} - \frac{3}{2} t^{2} + 8$ $v(t)=5/3t^3-3/2t^2+8$ in the interval $[0, 4]$ $[0,4]$

After that $t > 4$ $t>4$ , assume that the speed is constant

$v (4) = \frac{5}{3} \cdot 4^{3} - \frac{3}{2} \cdot 4^{2} + 8 = \frac{320}{3} - 28 + 8 = 86.67 m s^{- 1}$ $v(4)=5/3*4^3-3/2*4^2+8=320/3-28+8=86.67ms^-1$

The average speed on the interval $[0, 5]$ $[0,5]$ is

$5 ¯ v = \int_{0}^{4} (\frac{5}{3} t^{3} - \frac{3}{2} t^{2} + 8) d t + 86.67 \cdot 1$ $5barv=int_0^4(5/3t^3-3/2t^2+8)dt+86.67*1$

$= {[\frac{5}{12} t^{4} - \frac{1}{2} t^{3} + 8 t]}_{0}^{4} + 86.67$ $=[5/12t^4-1/2t^3+8t]_0^4+86.67$

$= (\frac{5}{12} \cdot 4^{2} - \frac{4^{3}}{2} + 8 \cdot 4) + 86.67$ $=(5/12*4^2-4^3/2+8*4)+86.67$

$= 193.3$ $=193.3$

The average speed is

$¯ v = \frac{193.3}{5} = 38.67 m s^{- 1}$ $barv=193.3/5=38.67ms^-1$

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What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $8 \frac{m}{s}$ $8 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 t^{2} - 3 t$ $a(t) =5t^2-3t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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What is the average speed, on t∈[0,5]t in [0,5], of an object that is moving at 8ms8 m/s at t=0t=0 and accelerates at a rate of a(t)=5t2−3ta(t) =5t^2-3t on t∈[0,4]t in [0,4]?

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Explanation:

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What is the average speed, on $t \in [0, 5]$ $t in [0,5]$ , of an object that is moving at $8 \frac{m}{s}$ $8 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 t^{2} - 3 t$ $a(t) =5t^2-3t$ on $t \in [0, 4]$ $t in [0,4]$ ?