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Math  /  Calculus

QuestionThe velocity function is v(t)=−t2+6t−8v(t)=-t^{2}+6 t-8 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-2,6]. displacement = −120.67-120.67 □\square distance traveled = 92.67 □\square

Studdy Solution

STEP 1

What is this asking? We're looking at a particle zipping around, and its speed changes over time according to a specific formula.
We need to figure out how far it shifts from its starting point and its total journey during a certain timeframe. Watch out! Displacement and distance traveled are *not* the same!
Displacement is simply the difference between the starting and ending positions, while distance traveled is the total ground covered, regardless of direction.

STEP 2

1. Analyze the velocity function
2. Calculate the displacement
3. Calculate the distance traveled

STEP 3

Alright, let's break down this velocity function!
We've got v(t)=−t2+6t−8v(t) = -t^2 + 6t - 8.
This tells us how fast the particle is moving at any given time tt.

STEP 4

To find where the particle changes direction, we need to find when the velocity is zero.
So, we set v(t)=0v(t) = 0: −t2+6t−8=0-t^2 + 6t - 8 = 0 Multiplying by -1 to make it easier to factor: t2−6t+8=0t^2 - 6t + 8 = 0 Factoring this quadratic equation, we get: (t−2)(t−4)=0(t-2)(t-4) = 0 So, the velocity is zero at t=2t = \mathbf{2} and t=4t = \mathbf{4}.
These are the times when the particle might be changing direction!

STEP 5

Our time interval is from t=−2t = \mathbf{-2} to t=6t = \mathbf{6}.
So, we're interested in what happens at t=−2t = -2, t=2t=2, t=4t=4, and t=6t=6.

STEP 6

Displacement is simply the integral of velocity over the given time interval.
Let's calculate that: ∫−26v(t) dt=∫−26(−t2+6t−8) dt\int_{-2}^{6} v(t) \, dt = \int_{-2}^{6} (-t^2 + 6t - 8) \, dt =[−13t3+3t2−8t]−26= \left[ -\frac{1}{3}t^3 + 3t^2 - 8t \right]_{-2}^{6}=(−13(6)3+3(6)2−8(6))−(−13(−2)3+3(−2)2−8(−2))= \left( -\frac{1}{3}(\mathbf{6})^3 + 3(\mathbf{6})^2 - 8(\mathbf{6}) \right) - \left( -\frac{1}{3}(\mathbf{-2})^3 + 3(\mathbf{-2})^2 - 8(\mathbf{-2}) \right)=(−72+108−48)−(83+12+16)= (-72 + 108 - 48) - (\frac{8}{3} + 12 + 16)=(−12)−(83+28)=−12−83−28=−40−83=−1203−83=−1283= (-12) - (\frac{8}{3} + 28) = -12 - \frac{8}{3} - 28 = -40 - \frac{8}{3} = -\frac{120}{3} - \frac{8}{3} = -\frac{\mathbf{128}}{3}So, the displacement is −128/3-128/3, or approximately −42.67\mathbf{-42.67}.

STEP 7

Now for the distance traveled.
Since the particle changes direction, we need to break down the integral into parts where the velocity is positive and negative.

STEP 8

From t=−2t=-2 to t=2t=2, the velocity is negative.
The distance traveled during this time is: ∣∫−22v(t) dt∣=∣[−13t3+3t2−8t]−22∣=∣−403−883∣=∣−1283∣=1283 \left| \int_{-2}^{2} v(t) \, dt \right| = \left| \left[ -\frac{1}{3}t^3 + 3t^2 - 8t \right]_{-2}^{2} \right| = \left| -\frac{40}{3} - \frac{88}{3} \right| = \left| -\frac{128}{3} \right| = \frac{\mathbf{128}}{3}

STEP 9

From t=2t=2 to t=4t=4, the velocity is positive.
The distance traveled during this time is: ∣∫24v(t) dt∣=∣[−13t3+3t2−8t]24∣=∣−163−(−403)∣=∣243∣=8 \left| \int_{2}^{4} v(t) \, dt \right| = \left| \left[ -\frac{1}{3}t^3 + 3t^2 - 8t \right]_{2}^{4} \right| = \left| -\frac{16}{3} - (-\frac{40}{3}) \right| = \left| \frac{24}{3} \right| = \mathbf{8}

STEP 10

From t=4t=4 to t=6t=6, the velocity is negative.
The distance traveled during this time is: ∣∫46v(t) dt∣=∣[−13t3+3t2−8t]46∣=∣−12−(−163)∣=∣−363+163∣=∣−203∣=203 \left| \int_{4}^{6} v(t) \, dt \right| = \left| \left[ -\frac{1}{3}t^3 + 3t^2 - 8t \right]_{4}^{6} \right| = \left| -12 - (-\frac{16}{3}) \right| = \left| -\frac{36}{3} + \frac{16}{3} \right| = \left| -\frac{20}{3} \right| = \frac{\mathbf{20}}{3}

STEP 11

The total distance traveled is the sum of these absolute values: 1283+8+203=1483+243=1723≈57.33 \frac{128}{3} + 8 + \frac{20}{3} = \frac{148}{3} + \frac{24}{3} = \frac{\mathbf{172}}{3} \approx \mathbf{57.33}

STEP 12

The displacement is −128/3≈−42.67-128/3 \approx -42.67.
The distance traveled is 172/3≈57.33172/3 \approx 57.33.

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