Question

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in a time of (t) seconds, a particle moves a distance of (s) meters from its starting point, where (s = 6t^{2}+3).
(a) find the average velocity between (t = 1) and (t = 1 + h) if:
(i) (h = 0.1) (ii) (h = 0.01) (iii) (h = 0.001)
enter the exact answers.
(i) when (h = 0.1), the average velocity between (t = 1) and (t = 1 + h) is (\text{i}) m/sec.
(ii) when (h = 0.01), the average velocity between (t = 1) and (t = 1 + h) is (\text{i}) m/sec.
(iii) when (h = 0.001), the average velocity between (t = 1) and (t = 1 + h) is (\text{i}) m/sec.
(b) use your answers to part (a) to estimate the instantaneous velocity of the particle at time (t = 1).
round your estimate to the nearest integer.
the instantaneous velocity appears to be (\text{i}) m/sec.
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Explanation:

Step1: Recall average - velocity formula

The average velocity $v_{avg}$ over the interval $[a, b]$ is given by $v_{avg}=\frac{s(b)-s(a)}{b - a}$. Here, $a = 1$, $b=1 + h$, and $s(t)=6t^{2}+3$. So $s(1 + h)=6(1 + h)^{2}+3=6(1 + 2h+h^{2})+3=6 + 12h+6h^{2}+3=9 + 12h+6h^{2}$ and $s(1)=6\times1^{2}+3=6 + 3=9$. Then $v_{avg}=\frac{s(1 + h)-s(1)}{(1 + h)-1}=\frac{(9 + 12h+6h^{2})-9}{h}=\frac{12h+6h^{2}}{h}=12 + 6h$.

Step2: Calculate for $h = 0.1$

Substitute $h = 0.1$ into $v_{avg}=12 + 6h$. Then $v_{avg}=12+6\times0.1=12 + 0.6 = 12.6$.

Step3: Calculate for $h = 0.01$

Substitute $h = 0.01$ into $v_{avg}=12 + 6h$. Then $v_{avg}=12+6\times0.01=12 + 0.06 = 12.06$.

Step4: Calculate for $h = 0.001$

Substitute $h = 0.001$ into $v_{avg}=12 + 6h$. Then $v_{avg}=12+6\times0.001=12 + 0.006 = 12.006$.

Step5: Estimate instantaneous - velocity

As $h$ approaches $0$, the average velocity approaches the instantaneous velocity. Looking at the values of the average velocity for $h = 0.1$, $h = 0.01$, and $h = 0.001$ (which are $12.6$, $12.06$, and $12.006$ respectively), we can estimate the instantaneous velocity at $t = 1$ to be $12$.

Answer:

(i) $12.6$
(ii) $12.06$
(iii) $12.006$
(b) $12$