Question

At 2:00 PM, a car’s speedometer reads 30 mi/h. At 2:10 PM, it reads 50 mi/h. Show that at some time between 2:00 and 2:10, the acceleration is exactly \(120\,{\rm{mi/}}{{\rm{h}}^{\rm{2}}}\).

Short answer

It is proved that at some time between 2:00 and 2:10, the acceleration is exactly \(120\,{\rm{mi/}}{{\rm{h}}^{\rm{2}}}\).

Step-by-step solution

The mean value theorem

Let \(f\) be a function that is continuous on the closed interval \(\left( {a,b} \right)\) and differentiable on the open interval \(\left( {a,b} \right)\). Then there is a number \(c \in \left( {a,b} \right)\) such as that \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( b \right)}}{{b - a}}\).

The acceleration of the object

The derivative of the velocity at any time is referred to as theacceleration of the object.

It is given that at 2:00 PM, a car’s speedometer reads 30 mi/h, and at 2:10 PM, it reads 50 mi/h. That means \(v\left( {2:00} \right) = 30\) and \(v\left( {2:10} \right) = 50\). By using Mean Value Theorem in the interval \(\left( {0,\frac{1}{6}} \right)\), there exists \(c\) such that:

\(\begin{aligned}{c}v'\left( c \right) &= \frac{{v\left( {2:10} \right) - v\left( {2:00} \right)}}{{\frac{1}{6} - 0}}\\a\left( c \right) &= \frac{{50 - 30}}{{\frac{1}{6}}}\\a\left( c \right) &= 120\end{aligned}\)

Where \(c\) is any point between \(2:00\) and \(2:10\) PM.

Hence, it is proved that at some time between 2:00 and 2:10 the acceleration is exactly \(120\,{\rm{mi/}}{{\rm{h}}^{\rm{2}}}\).