karush
Gold Member
MHB
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At $$ 2:00\text{ pm}$$ a car’s speedometer reads $$30\text { mph}$$.
At $$2:10$$ it reads $$50 \text { mph}$$.
Show that at some time between $$2:00$$ and $$2:10$$
the acceleration is exactly $$120\text{ mi} / h^2$$
This is in section of the Mean Value Theorem so since
$$
f'(c)=\frac{f(b)-f(a)}{b-a}
\Rightarrow
\frac{50-30}{10-0}
$$
I don't see that $120 mi/ h^2$ is going to fit into this also why is there $h^2$
At $$2:10$$ it reads $$50 \text { mph}$$.
Show that at some time between $$2:00$$ and $$2:10$$
the acceleration is exactly $$120\text{ mi} / h^2$$
This is in section of the Mean Value Theorem so since
$$
f'(c)=\frac{f(b)-f(a)}{b-a}
\Rightarrow
\frac{50-30}{10-0}
$$
I don't see that $120 mi/ h^2$ is going to fit into this also why is there $h^2$