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 $a (t) = 7 - t$ $a(t) =7-t$ 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 $a (t) = 7 - t$ $a(t) =7-t$ on $t \in [0, 4]$ $t in [0,4]$ ?

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Answer 1 · 2016-03-03T01:16:52

I found $5 \frac{m}{s}$ $5m/s$

Explanation:

I do not want to complicate it too much but I would integrate the acceleration to go back to the expression of velocity:
$v (t) = \int a (t) d t = \int (7 - t) d t = 7 t - \frac{t^{2}}{2} + c$ $v(t)=inta(t)dt=int(7-t)dt=7t-t^2/2+c$
now we get the constant $c$ $c$ by imposing that at $t = 0$ $t=0$ we have $v (0) = 12 \frac{m}{s}$ $v(0)=12m/s$ . So:
$12 = (7 \cdot 0) - \frac{{(0)}^{2}}{2} + c$ $12=(7*0)-(0)^2/2+c$
so that $c = 12 \frac{m}{s}$ $c=12m/s$
so the velocity will be:
$v (t) = 7 t - \frac{t^{2}}{2} + 12$ $v(t)=7t-t^2/2+12$
that at $t = 4 s$ $t=4s$ is:
$v (4) = (7 \cdot 4) - \frac{4^{2}}{2} + 12 = 32 \frac{m}{s}$ $v(4)=(7*4)-(4^2)/2+12=32m/s$

Average speed will be: $(Deltav)/(Deltat)=(v(4)-v(0))/(4-0)=(32-12)/4=5m/s$

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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 $a (t) = 7 - t$ $a(t) =7-t$ 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) =7-ta(t)=7−ta(t) =7-t 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 $a (t) = 7 - t$ $a(t) =7-t$ on $t \in [0, 4]$ $t in [0,4]$ ?