**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 $120 m/h^2$