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

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Answer 1 · 2018-04-03T06:29:29

The average speed is $= 13.25 m s^{- 1}$ $=13.25ms^-1$

Explanation:

The speed is the integral of the acceleration.

$a (t) = t^{2} - 2 t + 4$ $a(t)=t^2-2t+4$

$v (t) = \int (t^{2} - 2 t + 4) d t = \frac{1}{3} t^{3} - t^{2} + 4 t + C$ $v(t)=int(t^2-2t+4)dt=1/3t^3-t^2+4t+C$

Plugging in the initial conditions

$v (0) = 8$ $v(0)=8$

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

Therefore,

$v (t) = \frac{1}{3} t^{3} - t^{2} + 4 t + 8$ $v(t)=1/3t^3-t^2+4t+8$

The average speed is

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

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

$= \frac{1}{3} (\frac{81}{12} - 9 + 18 + 24)$ $=1/3(81/12-9+18+24)$

$= 13.25 m s^{- 1}$ $=13.25ms^-1$

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What is the average speed 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 $a (t) = t^{2} - 2 t + 4$ $a(t) =t^2-2t+4$ on $t \in [0, 3]$ $t in [0,3]$ ?

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Explanation:

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

1 Answer

Explanation:

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What is the average speed 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 $a (t) = t^{2} - 2 t + 4$ $a(t) =t^2-2t+4$ on $t \in [0, 3]$ $t in [0,3]$ ?