What is the average speed of an object that is still at $t=0$ and accelerates at a rate of $a(t) =t+3$ from $t in [2, 4]$ ?

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What is the average speed of an object that is still at $t=0$ and accelerates at a rate of $a(t) =t+3$ from $t in [2, 4]$ ?

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Answer 1 · 2016-03-04T12:53:58

Use the definition of acceleration and know that with respect to time, $u(0)=0$ because it is still. Also, you should give units of measurement (e.g. $m/s$ ). I didn't use any because you didn't give me.

$u_(aver)=14$

Explanation:

Being still at $t=0$ means that for $u=f(t)->u(0)=0$
Starting from the acceleration definition:

$a=(du)/dt$

$t+3=(du)/dt$

$(t+3)dt=du$

$int_0^t(t+3)dt=int_0^udu$

$int_0^(t)tdt+int_0^t3dt=int_0^udu$

$[t^2/2]_0^t+3[t]_0^t=[u]_0^u$

$(t^2/2-0^2/2)+3(t-0)=u-0$

$u(t)=t^2/2+3t$

So the average velocity between times 2 and 4 is:

$u_(aver)=(u(2)+u(4))/2$

$u(2)=2^2/2+3*2=8$

$u(4)=4^2/2+3*4=20$

Finally:

$u_(aver)=(8+20)/2$

$u_(aver)=14$

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What is the average speed of an object that is still at $t=0$ and accelerates at a rate of $a(t) =t+3$ from $t in [2, 4]$ ?

1 Answer

Explanation:

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Science

Math

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) =t+3a(t) =t+3 from t in [2, 4]t in [2, 4]?

1 Answer

Explanation:

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What is the average speed of an object that is still at $t=0$ and accelerates at a rate of $a(t) =t+3$ from $t in [2, 4]$ ?