What is the average speed of an object that is still at $t = 0$ $t=0$ and accelerates at a rate of $a (t) = 5 t + 3$ $a(t) =5t+3$ 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 $a (t) = 5 t + 3$ $a(t) =5t+3$ from $t \in [0, 2]$ $t in [0, 2]$ ?

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Answer 1 · 2016-04-22T09:34:17

$I found : v_{a} = 8$ $"I found :"v_a=8$

Explanation:

$The average value of a function for an interval of (a,b) is given as;$ $"The average value of a function for an interval of (a,b) is given as;"$
$v_{a} = \frac{1}{b - a} \int_{a}^{b} a (t) d t$ $v_a=1/(b-a)int _a^b a(t) d t$

$v_{a} = \frac{1}{2 - 0} \int_{0}^{2} (5 t + 3) d t$ $v_a=1/(2-0)int_0^2(5t+3)d t$

$v_{a} = \frac{1}{2} [{∣ ∣ ∣ 5 \frac{t^{2}}{2} + 3 t ∣ ∣ ∣}_{0}^{2}]$ $v_a=1/2[|5t^2/2+3t|_0^2]$

$v_{a} = \frac{1}{2} [(5 \cdot \frac{2^{2}}{2} + 3 \cdot 2) - (5 \cdot \frac{0^{2}}{2} + 3 \cdot 0)]$ $v_a=1/2[(5*2^2/2+3*2)-(5*0^2/2+3*0)]$

$v_{a} = \frac{1}{2} [(10 + 6) - 0]$ $v_a=1/2[(10+6)-0]$

$v_{a} = \frac{1}{2} \cdot 16$ $v_a=1/2*16$

$v_{a} = 8$ $v_a=8$

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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 $a (t) = 5 t + 3$ $a(t) =5t+3$ from $t \in [0, 2]$ $t in [0, 2]$ ?

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Humanities

... and beyond

What is the average speed of an object that is still at t=0t=0 and accelerates at a rate of a(t)=5t+3a(t) =5t+3 from t∈[0,2]t in [0, 2]?

1 Answer

Explanation:

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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 $a (t) = 5 t + 3$ $a(t) =5t+3$ from $t \in [0, 2]$ $t in [0, 2]$ ?