1. Particle Motion: Total Distance, Speeding Up and Slowing Down THOMAS DUNCAN

2. Background  You have to use the distance, velocity, and acceleration functions when finding a particle’s total distance travelled and when it is speeding up and slowing down. Therefore you must use derivatives with t as reference to time.  Assume the following:  F(x) is s(t) – the distance that the particle travelled.  F’(x) is v(t) – the velocity of the particle.  F’’(x) is a(t) – the acceleration of the particle.

3. Finding Total Distance Travelled  Some Calculus problems include calculating the total distance a particle travelled within a certain interval.  To find the total distance a particle travelled within an interval [a, b], subtract the distance between the two points and take the absolute value of that distance.  NOTE: Make sure whether or not a turning point is in the interval. Assuming that T is the turning point, you must do the following: |S(a) – S(T)| + |S(T) – S(b)|  This ensures that you actually calculated the entire distance the particle travelled.  For example: o Find the total distance a particle travelled from [0, 4] when S(0) = 2, S(3) = 5, and S(4) = 3. (t = 3 is a turning point) o Total distance = |2 – 5| + |5 – 3| = |-3| + |2| = 3 + 2 = 5

4. Speeding Up and Slowing Down  When the particle is speeding up, v(t) and a(t) have the same sign. (+,+ or –,–)  When the particle is slowing down, v(t) and a(t) have opposite signs (+,–)  You can determine this by finding the zeroes of v(t) and a(t) and finding whether values are + or – on a number line.  For example: o v(t) = 2t 2 + 5t and a(t) = 4t + 5  v(t) = t(2t + 5) and a(t) = 4t + 5 Therefore s(t) is slowing down at (-∞,-5/2) (-5/4, 0) and s(t) speeds up at (-5/2, -5/4) (0, ∞)

5. Example #1 s (t) = 3t 3 – 4t 2 + 1 v(t) = 9t 2 – 8t a(t) = 18t – 8 A) Find the total distance travelled from [0, 3]  Find the zeroes for v(t).  0 = t(9t – 8), t = 8/9, 0  Plug in the values for each point into s(t).  s(0) = 1 s(8/9) = -0.0535 s(3) = 46  Add and subtract: |1 – 0.0535| + |46 – (-0.0535)| = 0.9465 + 46.0535 = 47.000 B) Find where s(t) is speeding up and slowing down Speeding up: (0, 4/9) (8/9, ∞) Slowing down: (-∞, 0) (4/9, 8/9)

6. Example #2 s(t) = 5t 3 + 3t 2 + 2 v(t) = 15t 2 + 6t a(t) = 30t + 6 A) Find the total distance the particle travelled from [0, 2]  There are no turning points in this interval because v(t) factored has no positive zeroes.  Plug in values to s(t)  s(0) = 2 s(2) = 54  Subtract the difference: |54 – 2| = |52| = 52 B) Find where s(t) is speeding up and slowing down. Speeding Up: (-2/5, -1/5) (0, ∞) Slowing Down: (-∞, -2/5) (-1/5, 0)

7. Thank You