# Ap Question 2003

• March 22nd 2009, 06:10 PM
darleneflorio
Ap Question 2003
A particle moves along the x-axis so that its velocity at time t is given by v(t)=-(t+1)sin(t^2/2). During the time interval 0<t<3 or equal to t, what is the total distance between the particle and the origin? At t = 0 , the particle is at position x = 1.

The answer is x(0)+ antideriviative of v(t) from 0 to the square of 2pi.

x(0) must be where you start from and the total distance traveled is 4.334. But I'm still confused on why the distance is 2.265.
• March 22nd 2009, 06:29 PM
skeeter
Quote:

Originally Posted by darleneflorio
A particle moves along the x-axis so that its velocity at time t is given by v(t)=-(t+1)sin(t^2/2). During the time interval 0<t<3 or equal to t, what is the total distance between the particle and the origin? At t = 0 , the particle is at position x = 1.

The answer is x(0)+ antideriviative of v(t) from 0 to the square of 2pi.

x(0) must be where you start from and the total distance traveled is 4.334. But I'm still confused on why the distance is 2.265.

total distance traveled is not the same as distance from the origin.

if you walk 10 miles away from your house (the origin) and walk 9 miles back, the total distance you traveled is 19 miles ... that is not the distance you end up from your house, is it?

distance from the origin is $x(3) - 0 = x(3)$

$x(3) = x(1) + \int_1^3 v(t) \, dt$

because the FTC says ...

$x(3) - x(1) = \int_1^3 v(t) \, dt$