The velocity function is v(t) = - t^2 + 4 t - 3 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-2,6].

displacement = integral v dt

= -t^3/3 + 2 t^2 - 3 t
from -2 to +6
do the arithmetic

distance traveled
that means always use absolute value of any part of the integral
when is velocity negative and when is it positive?
0 = t^2 - 4 t + 3
0 = (t-3)(t-1)
velocity 0 at t = 1 and t = 3
so:
take absolute value of integral from
-2 to + 1
add to absolute value of integral from +1 to +3
add to absolute value of integral from +3 to + 6

thank you

To find the displacement, we need to integrate the velocity function over the given time interval. Let's start by finding the antiderivative of the velocity function.

Step 1: Find the antiderivative of v(t):
∫ (-t^2 + 4t - 3) dt

To integrate each term separately, use the power rule of integration:
∫ -t^2 dt + ∫ 4t dt - ∫ 3 dt

Using the power rule, we get:
= (-t^3/3) + (2t^2) - (3t) + C

Step 2: Evaluate the antiderivative at the upper and lower limits:
S(t) = (-t^3/3) + (2t^2) - (3t) + C

S(6) = [-(6^3)/3] + 2(6^2) - 3(6) + C
= -72 + 72 - 18 + C
= -18 + C

S(-2) = [-(2^3)/3] + 2(2^2) - 3(2) + C
= -8/3 + 8 - 6 + C
= 8/3 + C

Step 3: Calculate the displacement:
Displacement = S(6) - S(-2)
= (-18 + C) - (8/3 + C)
= -18 + C - 8/3 - C
= -18 - 8/3
= -54/3 - 8/3
= -62/3

Therefore, the displacement of the particle during the time interval [-2,6] is -62/3 units.

Now let's find the distance traveled by the particle.

Step 4: Calculate the absolute value of the displacement:
Distance = |Displacement|
= |-62/3|
= 62/3

Therefore, the distance traveled by the particle during the time interval [-2,6] is 62/3 units.

To find the displacement and distance traveled by the particle during the time interval [-2, 6], we need to integrate the velocity function over that interval. Here's how you can do it step by step:

1. Start with the given velocity function: v(t) = - t^2 + 4t - 3

2. To find the displacement, we need to integrate the velocity function with respect to time over the given interval [-2, 6]. The integral of a function represents the area under the curve of that function.

3. Integrate the velocity function v(t) over the interval [-2, 6]:
∫(v(t)) dt = ∫(-t^2 + 4t - 3) dt

4. Evaluate the definite integral by substituting the upper and lower limits of the interval:
∫(-t^2 + 4t - 3) dt = [-(1/3)t^3 + 2t^2 - 3t] from -2 to 6

5. Substitute the upper limit (6) and lower limit (-2), and subtract the result at the lower limit from the result at the upper limit:
Result = (-(1/3)(6)^3 + 2(6)^2 - 3(6)) - (-(1/3)(-2)^3 + 2(-2)^2 - 3(-2))

6. Simplify the expression:
Result = (-(1/3)(216) + 2(36) - 18) - (-(1/3)(-8) + 2(4) + 6)

7. Calculate the numerical value to get the final result.

To find the distance traveled by the particle during the time interval [-2, 6], we need to consider the absolute value of the velocity function and integrate it over that interval.

8. Write down the absolute value of the velocity function:
|v(t)| = |(-t^2 + 4t - 3)|

9. We need to break the interval [-2, 6] into smaller intervals based on the points where the velocity changes sign (from negative to positive or from positive to negative).

10. Solve the equation -t^2 + 4t - 3 = 0 to find the points where the velocity function changes sign.

11. Once you have the points where the velocity changes sign, break the interval [-2, 6] into smaller intervals accordingly.

12. For each smaller interval, integrate the absolute value of the velocity function over that interval.

13. Add up the individual integrals to find the total distance traveled.

Following these steps, you can find the displacement and distance traveled by the particle during the time interval [-2, 6].