Thread: 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.

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, be the velocity at and given time.
So on the closed interval the intivial point is,

The terminal point is,

Thus, the mean change is,

Hence, (assuming vecolity function is differenciable),
By Lagrange's Theorem (Mean Value),
There exists a point
Such as,