What is the average speed of an object that is moving at $22 \frac{m}{s}$ $22 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t - 8$ $a(t) =t-8$ on $t \in [0, 2]$ $t in [0,2]$ ?

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What is the average speed of an object that is moving at $22 \frac{m}{s}$ $22 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t - 8$ $a(t) =t-8$ on $t \in [0, 2]$ $t in [0,2]$ ?

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Answer 1 · 2018-03-17T09:35:17

The average speed is $= 3.67 m s^{- 1}$ $=3.67ms^-1$

Explanation:

$Average speed = \frac{total distance travelled}{time taken to cover that distance}$ $"Average speed " = "total distance travelled" /" time taken to cover that distance"$

The acceleration is

$a (t) = t - 8$ $a(t)=t-8$

The velocity is

$v (t) = \int a (t) d t = \int (t - 8) d t = \frac{1}{2} t^{2} - 8 t + C$ $v(t)=inta(t)dt = int(t-8)dt = 1/2t^2-8t+C$

Plugging in the initial conditions

$v (0) = 22 m s^{- 1}$ $v(0)=22ms^-1$

$22 = 0 - 0 + C$ $22=0-0+C$

Therefore,

$v (t) = \frac{1}{2} t^{2} - 8 t + 22$ $v(t)=1/2t^2-8t+22$

The average speed is

$¯ v = \frac{1}{2 - 0} \int_{0}^{2} (\frac{1}{2} t^{2} - 8 t + 22) d t$ $barv=1/(2-0)int_0^2(1/2t^2-8t+22)dt$

$= \frac{1}{2} {[\frac{1}{6} t^{3} - 4 t^{2} + 22 t]}_{0}^{3}$ $=1/2[1/6t^3-4t^2+22t]_0^2$

$= \frac{1}{2} ((\frac{1}{6} \cdot 8 - 16 + 22) - (0))$ $=1/2((1/6*8-16+22)-(0))$

$= 3.67 m s^{- 1}$ $=3.67ms^-1$

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What is the average speed of an object that is moving at $22 \frac{m}{s}$ $22 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t - 8$ $a(t) =t-8$ on $t \in [0, 2]$ $t in [0,2]$ ?

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What is the average speed of an object that is moving at 22ms22 m/s at t=0t=0 and accelerates at a rate of a(t)=t−8a(t) =t-8 on t∈[0,2]t in [0,2]?

1 Answer

Explanation:

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What is the average speed of an object that is moving at $22 \frac{m}{s}$ $22 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = t - 8$ $a(t) =t-8$ on $t \in [0, 2]$ $t in [0,2]$ ?