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At $2.00$ pm a car's speedometer reads $30$ miles/hr, at $2.10$ pm it reads $50$ miles/hr. Show that sometime between $2.00$ and $2.10$ the acceleration is exactly $120$ miles/hr$^{2}$.

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Toolbox:
  • Lagrange's Mean Value Theorem :
  • Let $f(x) $ be a real valued function that satisfies the following conditions.
  • (i) $f(x)$ is continuous on the closed interval $[a,b]$
  • (ii) $f(x)$ is differentiable in the open interval $(a,b)$
  • (iii) $f(a)=f(b)$
  • Then there exists atleast one value $c \in (a,b)$ such that $f'(c)=0$
Step 1:
The time interval in hours correcsponding to the time $2.00$ (taken as oh) to $2.10\; $(taken as $ \large\frac{10}{60}=\frac{1}{6}$$h)$ is $[0,\large\frac{1}{6}]$.
The speed is a continuous function of t in the closed interval $[0,\large\frac{1}{6}]$ and is differentiable on the open interval $(0, \large\frac{1}{6})$
Also $v(0)=30 \;and\; v(\large\frac{1}{6})$$=50$
Step 2:
By Lagrange's Mean Value Theorem : there is at least one instance 'c' when acceleration
$v'(c)=\large\frac{v\bigg(\Large\frac{1}{6}\bigg)-v(0)}{\Large\frac{1}{6}-0}$
$\qquad=\large\frac{50-30}{\large\frac{1}{6}}$$=120$
$\therefore$ at some time between 2.00 and 2.10 the acceleration is $120 m/h^2$

 

answered Jul 25, 2013 by meena.p
 
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