# calculus application

• Oct 30th 2006, 10:32 AM
bobby77
calculus 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 PM
ThePerfectHacker
Quote:

Originally Posted by bobby77
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.

Let, $v(t)$ be the velocity at and given time.
So on the closed interval $[2,2.1666]$ the intivial point is,
$v(2)=30$
The terminal point is,
$v(2.1666)=50$
Thus, the mean change is,
$\frac{50-30}{\frac{1}{6}}=120$
Hence, (assuming vecolity function is differenciable),
By Lagrange's Theorem (Mean Value),
There exists a point $t\in (2,2.16666)$
Such as,
$v'(t)=a(t)=120$