Browse Questions

# 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$^{2}$.

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
$v'(c)=\large\frac{v\bigg(\Large\frac{1}{6}\bigg)-v(0)}{\Large\frac{1}{6}-0}$