A particle moves along the X axis with the acceleration a(t)=12t-6. If v(0)=3, find the total distance the particle traveled from t=0 to t=2.
Please help
a = 12t-6
v = 6t^2-6t+c
v(0)=3, so c=3, and
v = 6t^2-6t+3
s = 2t^3-3t^2+3t+c
Since we assume zero distance at t=0, c=0.
s(2) = 2*8 - 3*4 + 3*2 = 10
To find the total distance traveled by the particle from t=0 to t=2, we need to find the displacement and then take the absolute value.
First, we integrate the acceleration function to find the velocity function:
v(t) = ∫ a(t) dt = ∫ (12t - 6) dt
= 6t^2 - 6t + C
Next, we use the initial condition v(0) = 3 to find the constant C:
3 = 6(0)^2 - 6(0) + C
3 = 0 + 0 + C
C = 3
Now we have the velocity function:
v(t) = 6t^2 - 6t + 3
To find the displacement function, we integrate the velocity function:
s(t) = ∫ v(t) dt = ∫ (6t^2 - 6t + 3) dt
= 2t^3 - 3t^2 + 3t + D
Using the initial condition s(0) = 0 to find the constant D:
0 = 2(0)^3 - 3(0)^2 + 3(0) + D
0 = 0 - 0 + 0 + D
D = 0
Now we have the displacement function:
s(t) = 2t^3 - 3t^2 + 3t
To find the total distance traveled, we take the absolute value of the displacement function and integrate it over the given interval [0, 2]:
distance = ∫ |s(t)| dt from 0 to 2
= ∫ |2t^3 - 3t^2 + 3t| dt from 0 to 2
To solve this, we split the integral into two parts based on the different intervals where the function is negative or positive:
distance = ∫ (2t^3 - 3t^2 + 3t) dt from 0 to 1 + ∫ (-(2t^3 - 3t^2 + 3t)) dt from 1 to 2
Simplifying the integrals:
distance = ∫ (2t^3 - 3t^2 + 3t) dt from 0 to 1 - ∫ (2t^3 - 3t^2 + 3t) dt from 1 to 2
Calculating the integrals:
distance = [(1/2)t^4 - t^3 + (3/2)t^2] evaluated from 0 to 1 - [(1/2)t^4 - t^3 + (3/2)t^2] evaluated from 1 to 2
Evaluating these expressions:
distance = [(1/2)(1)^4 - (1)^3 + (3/2)(1)^2] - [(1/2)(0)^4 - (0)^3 + (3/2)(0)^2] - [(1/2)(2)^4 - (2)^3 + (3/2)(2)^2] + [(1/2)(1)^4 - (1)^3 + (3/2)(1)^2]
Simplifying:
distance = [1/2 - 1 + 3/2] - [0 - 0 + 0] - [8/2 - 8 + 6] + [1/2 - 1 + 3/2]
= [3/2] - [14] + [3/2]
= 3/2 - 14 + 3/2
= 12/2 - 14
= 6 - 14
= -8
Since distance cannot be negative, we take the absolute value:
distance = |-8| = 8
Therefore, the total distance traveled by the particle from t=0 to t=2 is 8 units.
To find the total distance traveled by the particle, we need to integrate the absolute value of the velocity function over the given time interval.
First, let's find the velocity function v(t) by integrating the acceleration function a(t) with respect to time:
v(t) = ∫ (12t - 6) dt
Using the power rule of integration:
v(t) = (1/2) * (12t^2) - 6t + C
Now, we can find the constant C using the given initial condition v(0) = 3:
3 = (1/2) * (12 * 0^2) - 6 * 0 + C
3 = C
Therefore, the velocity function is:
v(t) = (1/2) * (12t^2) - 6t + 3
Next, we need to find the total distance traveled. This can be done by integrating the absolute value of the velocity function over the given time interval:
Total distance = ∫ |v(t)| dt from t = 0 to t = 2
To calculate this integral, we need to consider two cases:
1. If v(t) is positive (or zero) over the time interval, the absolute value of v(t) is equal to v(t).
2. If v(t) is negative over the time interval, the absolute value of v(t) is equal to -v(t).
Let's break up the integral into these two cases:
Total distance = ∫ v(t) dt from t = 0 to t = 2 if v(t) ≥ 0
= -∫ v(t) dt from t = 0 to t = 2 if v(t) < 0
Now, let's evaluate the integral for both cases:
Case 1:
Total distance = ∫ [(1/2) * (12t^2) - 6t + 3] dt from t = 0 to t = 2
Integrating each term:
Total distance = (1/2) * [4t^3 - 3t^2 + 3t] from t = 0 to t = 2
Plugging in the limits:
Total distance = (1/2) * [(4*(2)^3 - 3*(2)^2 + 3*(2))] - [(1/2) * (4*(0)^3 - 3*(0)^2 + 3*(0))]
Total distance = (1/2) * [32 - 12 + 6] - [(1/2) * 0]
Total distance = (1/2) * [26] - 0
Total distance = 13 units
Case 2:
Total distance = -∫ [(1/2) * (12t^2) - 6t + 3] dt from t = 0 to t = 2
Integrating each term:
Total distance = - (1/2) * [4t^3 - 3t^2 + 3t] from t = 0 to t = 2
Plugging in the limits:
Total distance = - (1/2) * [(4*(2)^3 - 3*(2)^2 + 3*(2))] - [-(1/2) * (4*(0)^3 - 3*(0)^2 + 3*(0))]
Total distance = - (1/2) * [32 - 12 + 6] - [- (1/2) * 0]
Total distance = - (1/2) * [26] + 0
Total distance = -13 units
Now, since the displacement is a vector quantity indicating direction, we can conclude that the total distance traveled by the particle from t=0 to t=2 is 13 units.