Particle Motion: Total Distance, Speeding Up and Slowing DownTHOMAS DUNCAN

BackgroundYou 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.

Finding Total Distance TravelledSome 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:

oFind 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)

oTotal distance = |2 – 5| + |5 – 3| = |-3| + |2| = 3 + 2 = 5

Speeding Up and Slowing DownWhen 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:

ov(t) = 2t2 + 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, ∞)

Example #1•s (t) = 3t3 – 4t2 + 1 v(t) = 9t2 – 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)

Example #2• s(t) = 5t3 + 3t2 + 2 v(t) = 15t2 + 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)

Thank You