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Math  /  Calculus

QuestionSuppose that a particle moves along a straight line with a velocity v(t)=4−2tv(t)=4-2 t, where tt is in the interval [0,8][0,8]. Find the displacement of the particle up to t=8t=8 and the total distance traveled up to t=8t=8.
Total displacement == □\square Total distance == □\square

Studdy Solution

STEP 1

What is this asking? We're looking at a particle zipping along a line, and we want to know how far it's moved from its starting point after 8 seconds, and also its total travel distance regardless of direction. Watch out! Displacement can be tricky!
It's not always the same as the total distance traveled.
Think about it – if you walk forward three steps and then back three steps, your displacement is zero, but you've walked six steps total!

STEP 2

1. Find when the particle changes direction
2. Calculate the displacement
3. Calculate the total distance

STEP 3

To find when the particle changes direction, we need to find when its velocity is **zero**.
This is because the particle changes direction when it momentarily stops!

STEP 4

So, we set v(t)=0v(t) = 0 and solve for tt: 4−2t=04 - 2t = 0 4=2t4 = 2tt=42t = \frac{4}{2}t=2t = \textbf{2}This means the particle changes direction at t=2t = \textbf{2} seconds.

STEP 5

Displacement is simply the integral of velocity.
It represents the **net change in position** from the starting point.

STEP 6

We calculate the displacement from t=0t = 0 to t=8t = 8 by integrating the velocity function: ∫08v(t) dt=∫08(4−2t) dt \int_{0}^{8} v(t) \, dt = \int_{0}^{8} (4 - 2t) \, dt =[4t−t2]08 = \left[ 4t - t^2 \right]_0^8 =(4⋅8−82)−(4⋅0−02) = (4 \cdot \textbf{8} - \textbf{8}^2) - (4 \cdot \textbf{0} - \textbf{0}^2) =(32−64)−(0−0) = (32 - 64) - (0 - 0) =−32 = -32 So, the displacement is -32\textbf{-32}.

STEP 7

The total distance is the sum of the distances traveled in each direction.
We know the particle changes direction at t=2t = \textbf{2}, so we'll calculate the distance traveled from t=0t = 0 to t=2t = \textbf{2} and from t=2t = \textbf{2} to t=8t = 8, and then add the absolute values of these distances together.

STEP 8

Distance from t=0t = 0 to t=2t = \textbf{2}: ∫02(4−2t) dt=[4t−t2]02 \int_{0}^{2} (4 - 2t) \, dt = \left[ 4t - t^2 \right]_0^2 =(4⋅2−22)−(4⋅0−02) = (4 \cdot \textbf{2} - \textbf{2}^2) - (4 \cdot \textbf{0} - \textbf{0}^2) =(8−4)−(0−0) = (8 - 4) - (0 - 0) =4 = 4

STEP 9

Distance from t=2t = \textbf{2} to t=8t = 8: ∫28(4−2t) dt=[4t−t2]28 \int_{2}^{8} (4 - 2t) \, dt = \left[ 4t - t^2 \right]_2^8 =(4⋅8−82)−(4⋅2−22) = (4 \cdot \textbf{8} - \textbf{8}^2) - (4 \cdot \textbf{2} - \textbf{2}^2) =(32−64)−(8−4) = (32 - 64) - (8 - 4) =−32−4 = -32 - 4 =−36 = -36

STEP 10

Now, we add the absolute values of these distances: ∣4∣+∣−36∣=4+36=40 |4| + |-36| = 4 + 36 = \textbf{40} So, the total distance traveled is 40\textbf{40}.

STEP 11

Total displacement = −32-32 Total distance = 4040

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