At 2.00pm the speedometer of your car read 30km/hr.At 2.10pm it reads 50km/hr.show that at sometime between 2.00pm and 2.10pm the acceleration is exactly 120km/hr^2.

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- Oct 30th 2006, 10:32 AMbobby77calculus application
At 2.00pm the speedometer of your car read 30km/hr.At 2.10pm it reads 50km/hr.show that at sometime between 2.00pm and 2.10pm the acceleration is exactly 120km/hr^2.

- Oct 30th 2006, 12:06 PMThePerfectHacker
Let, $\displaystyle v(t)$ be the velocity at and given time.

So on the closed interval $\displaystyle [2,2.1666]$ the intivial point is,

$\displaystyle v(2)=30$

The terminal point is,

$\displaystyle v(2.1666)=50$

Thus, the mean change is,

$\displaystyle \frac{50-30}{\frac{1}{6}}=120$

Hence, (assuming vecolity function is differenciable),

By Lagrange's Theorem (Mean Value),

There exists a point $\displaystyle t\in (2,2.16666)$

Such as,

$\displaystyle v'(t)=a(t)=120$