Math  /  Calculus

Question1. (6402#3\left(6402 \# 3\right. p. 303) A particle is moving on the xx-axis with velocity given by v(l)=3ι2+18ι15v(l)=-3 \iota^{2}+18 \iota-15. Find both the displacement (net change of position) and the total distance travelled for this particle during the time interval 0ι80 \leq \iota \leq 8.

Studdy Solution

STEP 1

What is this asking? We need to figure out how far a particle *moves* on a line, and also the *total ground* it covers, between time 0 and time 8. Watch out! Displacement and total distance are *not* the same!
Displacement is how far the particle ends up from where it started.
Total distance is like tracking the mileage on your car – it only goes up!

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 its velocity switches from positive to negative or vice-versa.
So, let's set v(t)v(t) equal to zero and solve for tt:
3t2+18t15=0-3t^2 + 18t - 15 = 0

STEP 4

We can **divide** both sides by 3-3 to simplify:
t26t+5=0t^2 - 6t + 5 = 0

STEP 5

Now, we can **factor** the quadratic:
(t1)(t5)=0(t - 1)(t - 5) = 0

STEP 6

This gives us two **times** when the velocity is zero: t=1t = 1 and t=5t = 5.
These are our critical times!

STEP 7

The displacement is simply the **integral** of the velocity from the **start time** (t=0t = 0) to the **end time** (t=8t = 8).
Let's calculate that:
08(3t2+18t15)dt\int_{0}^{8} (-3t^2 + 18t - 15) \, dt

STEP 8

**Calculate the integral**:
[t3+9t215t]08\left[ -t^3 + 9t^2 - 15t \right]_0^8

STEP 9

**Plug in** the limits of integration:
((8)3+9(8)215(8))((0)3+9(0)215(0))(-(8)^3 + 9(8)^2 - 15(8)) - (-(0)^3 + 9(0)^2 - 15(0))(512+576120)(0)=56(-512 + 576 - 120) - (0) = -56

STEP 10

So, the displacement is 56-56.
This means the particle ends up **56 units to the left** of where it started.

STEP 11

To find the total distance, we need to take the **absolute value** of the integral between each change of direction.
We already found those changes happen at t=1t = 1 and t=5t = 5.
So, we need to calculate the integral between 00 and 11, between 11 and 55, and between 55 and 88, and take the absolute value of each.

STEP 12

**Calculate** the first integral (from 0 to 1):
01(3t2+18t15)dt=13+9(12)15(1)0=7=7\left| \int_{0}^{1} (-3t^2 + 18t - 15) \, dt \right| = |-1^3 + 9(1^2) - 15(1) - 0| = |-7| = 7

STEP 13

**Calculate** the second integral (from 1 to 5):
15(3t2+18t15)dt=53+9(52)15(5)(1+915)=125+22575(7)=32=32\left| \int_{1}^{5} (-3t^2 + 18t - 15) \, dt \right| = |-5^3 + 9(5^2) - 15(5) - (-1 + 9 - 15)| = |-125 + 225 - 75 - (-7)| = |32| = 32

STEP 14

**Calculate** the third integral (from 5 to 8):
58(3t2+18t15)dt=83+9(82)15(8)(53+9(52)15(5))=5625=88=88\left| \int_{5}^{8} (-3t^2 + 18t - 15) \, dt \right| = |-8^3 + 9(8^2) - 15(8) - (-5^3 + 9(5^2) - 15(5))| = |-56 - 25| = |-88| = 88

STEP 15

**Add up** the absolute values to get the total distance:
7+32+88=1277 + 32 + 88 = 127

STEP 16

The displacement of the particle is 56-56, and the total distance traveled is 127127.

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