The velocity function is v(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 [-1,6].
v=ds/dt
ds/dt=t^2-6t+8
ds=(t^2-6t+8)*dt
s=integral(t^2-6t+8)*dt
s=(t^3-9t^2+24t)/3+C
s=(t/3)*(t^2-9t+24)+C
s=(t/3)*[(t-9)*t+24)]+C
s=Definite integral betwen(-1,6)
s=(6/3)*[(6-9)*6+24]-(-1/3)*[(-1-9)*(-1)+24]
s=(6/3)*[(-3)*6+24]+(1/3)*[-10*(-1)+24]
s=2*(-18+24)+(1/3)*(10+24)
s=2*6+(1/3)*34
s=12+34/3
s=(36/3)+(34/3)
s=70/3
To find the displacement and distance traveled by the particle during the time interval [-1,6], we first need to integrate the velocity function.
The displacement of the particle can be found by taking the definite integral of the velocity function over the given time interval. The integral of v(t) from -1 to 6 will give us the change in position or displacement.
To find the definite integral, we integrate each term of the velocity function separately and then evaluate it over the interval [-1,6].
∫(t^2 - 6t + 8)dt = ∫t^2 dt - ∫6t dt + ∫8 dt
Integrating each term separately:
∫t^2 dt = (t^3)/3
∫6t dt = 3t^2
∫8 dt = 8t
Now, we evaluate each integral from -1 to 6:
Displacement = [(6^3)/3 - (-1^3)/3] - [3(6^2) -3(-1^2)] + [8(6) - 8(-1)]
Simplifying this expression, we get:
Displacement = [(216/3) - (-1/3)] - [3(36) - 3] + [48 + 8]
Displacement = [72 + 1] - [108 - 3] + [48 + 8]
Displacement = 73 - 105 + 56
Displacement = 24
Therefore, the particle has a displacement of 24 units during the time interval [-1,6].
To find the distance traveled, we need to consider the absolute value of the function. This is because velocity can be either positive or negative depending on the direction of the particle's motion.
So, we integrate the absolute value of the velocity function over the interval [-1,6]:
Distance = ∫|t^2 - 6t + 8| dt
To evaluate this integral, we need to find the critical points where the absolute value function changes sign. These occur where t^2 - 6t + 8 = 0.
Factoring the quadratic equation, we have: (t - 4)(t - 2) = 0
Therefore, the critical points are t = 4 and t = 2.
We then evaluate the integral over three separate intervals: [-1,2], [2,4], and [4,6].
For the interval [-1,2]:
Distance = ∫(2 - t^2 - 6t + 8) dt = ∫(t^2 + 6t - 2) dt = [(t^3)/3 + 3t^2 - 2t] evaluated from -1 to 2
Simplifying this expression, we get:
Distance = [(2^3)/3 + 3(2^2) - 2(2)] - [((-1)^3)/3 + 3((-1)^2) - 2(-1)]
Distance = [8/3 + 12 - 4] - [-1/3 + 3 + 2]
Distance = [8/3 + 12 - 4] - [-1/3 + 5]
Distance = [8/3 + 12 - 4] - [14/3]
Distance = [8 + 36 - 12] - [14]
Distance = 32 - 14
Distance = 18
For the interval [2,4]:
Distance = ∫(-t^2 + 6t - 8) dt = [(-t^3)/3 + 3t^2 - 8t] evaluated from 2 to 4
Simplifying this expression, we get:
Distance = [(-4^3)/3 + 3(4^2) - 8(4)] - [((-2)^3)/3 + 3((-2)^2) - 8(-2)]
Distance = [-64/3 + 48 - 32] - [-8/3 + 12 + 16]
Distance = [-64/3 + 16] - [20/3]
Distance = [-4/3] - [20/3]
Distance = -24/3
Distance = -8
For the interval [4,6]:
Distance = ∫(t^2 - 6t + 8) dt = [(t^3)/3 - 3t^2 + 8t] evaluated from 4 to 6
Simplifying this expression, we get:
Distance = [(6^3)/3 - 3(6^2) + 8(6)] - [(4^3)/3 - 3(4^2) + 8(4)]
Distance = [72/3 - 108 + 48] - [64/3 - 48 + 32]
Distance = [24 - 108 + 48] - [64/3 - 48 + 32]
Distance = [-36/3 + 72] - [48/3]
Distance = (72 - 36) - (16)
Distance = 36 - 16
Distance = 20
Finally, to find the total distance traveled, we add up the distances for each interval:
Total Distance = Distance[-1,2] + Distance[2,4] + Distance[4,6]
Total Distance = 18 + (-8) + 20
Total Distance = 30
Therefore, the particle traveled a distance of 30 units during the time interval [-1,6].