I am learning how to use the equation $\frac {1}{b-a} \int^b_af(x)dx$. More secifically I dont understand how to apply it to the following problem:
A car starting from rest accelerates at a rate of 2 m/sec2 . Find the average speed of this object in the first 10 seconds.
$\endgroup$2 Answers
$\begingroup$Given an integrable curve, say, for speed, between times $a$ and $b$, the average speed is pretty intuitive to grasp. If you take the speed for each second, add them up, and divide by the number of seconds, you get a good estimate. If you take the limit of this with smaller and smaller rectangles, you get the accurate value for average speed, and this is given by
$$\frac{1}{b-a}\int_a^b f(x)\ dx.$$
In this case, the acceleration is given by $a(t)=2$, so the velocity is simply $v(t)=2t+C$. Since the car starts from rest, the constant $C$ must be $0$, so we have the speed function $v(t)=2t$. Thus, in the first ten seconds, the above formula gives the average speed $A$:
$$A=\frac{1}{10-0}\int_0^{10}2t\ dt=\frac{1}{10}t^2\bigg|_0^{10}=\frac{1}{10}\left(10^2-0\right)=10.$$
$\endgroup$ 5 $\begingroup$At time $t$ his speed is $at$, where $a$ is the acceleration. Hence the average speed is $$ \frac{1}{t_0-0}\int_0^{t_0} at \mathrm{d}t=\frac{1}{t_0}\cdot \frac{1}{2}at_0^2=\frac{1}{2}at_0. $$
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