Rectilinear+Motion+Problems

=Rectilinear Motion= One of the first and simplest applications of differentiation is the analysis of objects in rectilinear motion, or motion is a straight line. You will be given an equation that relates the object's position, velocity, or acceleration as it relates to the time of travel, and will be asked to describe aspects of the object's motion.

Before learning integration in the next unit, these problems will almost always begin with a position function: s(t) that tells you the particles locations at a given time. So how can we describe the velocity or acceleration of the object? Notice that velocity is determined by distance divided by time, and that the position function's graph shows its position in relation to time. Therefore you can find the velocity by finding the slope of the curve. You can find the instantaneous velocity by taking the derivative of the position function.

Similarly, acceleration is in distance per unit of time squared. This is the slope of, or the derivative of, the velocity function.

A particle's position along the x-axis, in meters, is given by the function s(t)=(1/3)t 3 -(5/2)t 2 +6t+1, with time measured in seconds.
 * Example One **
 * A**. Find the velocity and acceleration at t=1 second. **B**. Find when the distance from the origin goes from increasing to decreasing.
 * C**. Find the net and total distance traveled by the particle on the time interval 0<t<12 seconds.

Solution: With these equations, we have v(1)=1-5+6=**2 m/sec** and a(1)=2-5=**(-3) m/s 2 **. We will first find the critical points of s(t) by setting v(t)=0: 0=(t-2)(t-3) yielding t=2, 3. a(2)=(-1) and a(3)=1, so **t=2 seconds** is our relative maximum by the Second Derivative Test. The total distance must account for the distance traveled as the particle changes direction, so we must add the distances traveled over the intervals defined by when the direction is changing. From Part B, these intervals are on [0,2] [2,3] and [3,12]. This yields: [s(2)-s(0)]+[s(2)-s(3)]+[s(12)-s(3)]= **732m**.
 * A**. The velocity function, v(t)=s'(t)=t 2 -5t+6 and the acceleration function, a(t)=v '(t)=2t-5.
 * B**. Position goes from increasing to decreasing when Velocity goes from positive to negative.
 * C**. The net distance traveled, or change in position, is simply the final position minus the initial position: s(12)-s(0)=289m - 1m= **288m**.