What is the average speed of an object that is moving at $12 \frac{m}{s}$ $12 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 2 - 5 t$ $a(t) =2-5t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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What is the average speed of an object that is moving at $12 \frac{m}{s}$ $12 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 2 - 5 t$ $a(t) =2-5t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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Answer 1 · 2018-03-15T14:51:14

Given, acceleration = $a = \frac{d v}{d t} = 2 - 5 t$ $a=(dv)/(dt)=2-5t$

so, $v = 2 t - \frac{5 t^{2}}{2} + 12$ $v=2t -(5t^2)/2 +12$ (by integration)

Hence, $v = \frac{d x}{d t} = 2 t - \frac{5 t^{2}}{2} + 12$ $v=(dx)/(dt)=2t-(5t^2)/2 +12$

so, $x = t^{2} - \frac{5}{6} t^{3} + 12 t$ $x=t^2 -5/6 t^3 +12t$

Putting, $x = 0$ $x=0$ we get, $t = 0, 3.23$ $t=0,3.23$

So,total distance covered = ${[t^{2}]}_{0}^{2} - \frac{5}{6} {[t^{3}]}_{0}^{3} + 12 {[t]}_{0}^{3.23} + \frac{5}{6} {[t^{3}]}_{3.23}^{3} - {[t^{2}]}_{3.23}^{2} - 12 {[t]}_{3.23}^{4} = 31.54 m$ $[t^2]_0^(3.23) -5/6 [t^3]_0^3.23 +12[t]_0^3.23+5/6[t^3]_3.23^4 -[t^2]_3.23^4 -12[t]_3.23^4=31.54m$

So,average velocity = total distance covered/total time taken = $\frac{31.54}{4} = 7.87 m s^{- 1}$ $31.54/4=7.87 ms^-1$

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What is the average speed of an object that is moving at $12 \frac{m}{s}$ $12 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 2 - 5 t$ $a(t) =2-5t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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What is the average speed of an object that is moving at 12 m/s12ms12 m/s at t=0t=0t=0 and accelerates at a rate of a(t) =2-5ta(t)=2−5ta(t) =2-5t on t in [0,4]t∈[0,4]t in [0,4]?

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What is the average speed of an object that is moving at $12 \frac{m}{s}$ $12 m/s$ at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 2 - 5 t$ $a(t) =2-5t$ on $t \in [0, 4]$ $t in [0,4]$ ?