APPLICATIONS OF INTEGRATION: MOTION ALONG A STRAIGHT LINE MR. VELAZQUEZ AP CALCULUS
UNDERSTANDING RECTILINEAR MOTION Rectilinear motion or motion along a straight line can now be understood in terms of differentiation and integration. If we suppose that the position of a particle along a straight line, relative to some fixed point, is given by a function of time s = F(t), then the velocity and acceleration of the particle can be represented thusly: v = ds dt a = dv = d2 s dt dt 2 We already explored the relationship between derivatives and motion in a previous section. In this section, we will be applying integration to deal will such problems.
DISTANCE TRAVELED BY A PARTICLE If the velocity of a particle is given as a function of time v = f(t), then the distance traveled by a particle from time t = a to t = b can be represented with the following integral: b න a f t dt Notice the presence of the absolute value symbol. This means we must take the negative areas included in the integral and make them positive. Remember the difference between displacement (the net change in the position of the particle relative to where it began, independent of path) and the distance (the total amount of movement experienced by the particle, in either direction).
Suppose a body moves along a straight line with velocity v = t 3 + 3t 2. What is the distance traveled by the particle between t = 1 and t =? Note from the graph that for all t > 0, the velocity is positive, or v > 0. This makes our job easier, as we do not have to consider any negative areas or split the integral in any way. b න a v t dt = න 1 t 3 + 3t 2 dt = t + t3 1 = 507
Suppose a body moves along the x-axis with velocity v t = 6t 2 18t + 12. (a) Find the total distance traveled by the particle between t = 0 and t =. (b)find the displacement of the particle from t = 0 to t = We don t necessarily need a graph for this one. First, we factor the velocity: v t = 6t 2 18t + 12 = 6 t 2 3t + 2 = 6(t 1)(t 2) This tells us that the particle changes direction at t = 1 and t = 2. We can also use a few test points to determine: Velocity is positive when t < 1 Velocity is negative when 1 < t < 2 Velocity is positive when t > 2 We can now define the total distance for part (a) of the problem: 1 න 0 v t dt න 1 2 v t dt + න 2 v t dt = 3
Suppose a body moves along the x-axis with velocity v t = 6t 2 18t + 12. (a) Find the total distance traveled by the particle between t = 0 and t =. (b)find the displacement of the particle from t = 0 to t = To find the displacement required for part (b), we simply apply FTC: න 0 v t dt = න 0 6t 2 18t + 12 dt = 2t 3 9t 2 + 12t 0 = 32 So while the total distance moved by the particle is 3 units, some of that motion was in the negative direction, meaning that the particle ends up with a net position change of 32 units.
A car moving with initial velocity of 5 mph accelerates at the rate of a t = 2. t mph per second for 8 seconds. (a) How fast is the car going when the 8 seconds are up? (b)how far did the car travel during those 8 seconds?
The acceleration of an object moving on a line is given at time t by a = sin t (at t = 0, the object is at rest). Find the distance the object travels from t = 0 to t = 5π 6.
(NET CONSUMPTION) From 1970 to 1980, the rate of potato consumption in a particular country was C t = 2. 2 + e 0.095 t millions of bushels per year, with t being the number of years (since the beginning of 1970). How many bushels were consumed from the start of 1972 to the end of 1973?
CLASSWORK Suppose a particle moves along the x-axis with velocity v = 3t 2 6t. At time t = 0, the particle is at a position x = 0 a) Derive the equations for the position, and acceleration of the particle at any time t b) Sketch the graphs for the position, velocity and acceleration of the particle from time t = 0 to t = 5. (1 graph for each) c) Find the total distance traveled by the particle from t = 0 to t = 3 d) Find the total displacement of the particle from t = 0 to t = 5 Homework: Khan Academy (Due 2/1)