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The velocity function (in meters per second) for a particle moving along a line is given by

v(t)=-2(t^2-2t-8), 1<=t<=6
(a) Find the displacement (in meters) of the particle.

(b) Find the total distance traveled (in meters) by the particle.

User Bdd
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Answer:

To find the displacement of the particle, we need to calculate the definite integral of the velocity function over the given time interval.

(a) Displacement:

The displacement of the particle is given by the definite integral of the velocity function from t = 1 to t = 6:

Displacement = ∫[1 to 6] v(t) dt

First, let's find the antiderivative of the velocity function:

∫(-2(t^2 - 2t - 8)) dt = -2∫(t^2 - 2t - 8) dt

Integrating each term separately:

= -2 * [ (1/3)t^3 - t^2 - 8t ] evaluated from t = 1 to t = 6

= -2 * [ (1/3)(6^3) - (6^2) - 8(6) - (1/3)(1^3) + (1^2) + 8(1) ]

= -2 * [ (216/3) - 36 - 48 - (1/3) + 1 + 8 ]

= -2 * [ 72 - 36 - 48 - 1 + 1 + 8 ]

= -2 * [ -4 ]

= 8 meters

Therefore, the displacement of the particle is 8 meters.

(b) Total distance traveled:

To find the total distance traveled by the particle, we need to consider the absolute value of the velocity function over the given time interval.

Total distance traveled = ∫[1 to 6] |v(t)| dt

First, let's find the absolute value of the velocity function:

|v(t)| = |-2(t^2 - 2t - 8)| = 2|t^2 - 2t - 8|

Now, let's find the integral of the absolute value of the velocity function:

∫[1 to 6] 2|t^2 - 2t - 8| dt

To evaluate this integral, we need to split the integral into two parts based on the intervals where the function inside the absolute value changes sign.

For t = 1 to t = 3:

∫[1 to 3] 2(t^2 - 2t - 8) dt = 2∫[1 to 3] (t^2 - 2t - 8) dt

Integrating each term separately:

= 2 * [ (1/3)t^3 - t^2 - 8t ] evaluated from t = 1 to t = 3

= 2 * [ (1/3)(3^3) - (3^2) - 8(3) - (1/3)(1^3) + (1^2) + 8(1) ]

= 2 * [ (27/3) - 9 - 24 - (1/3) + 1 + 8 ]

= 2 * [ 9 - 9 - 24 - 1 + 1 + 8 ]

= 2 * [ -16 ]

= -32 meters (taking into account the negative sign)

For t = 3 to t = 6:

∫[3 to 6] 2(t^2 - 2t - 8) dt = 2∫[3 to 6] (t^2 - 2t - 8) dt

Integrating each term separately:

= 2 * [ (1/3)t^3 - t^2 - 8t ] evaluated from t = 3 to t = 6

= 2 * [ (1/3)(6^3) - (6^2) - 8(6) - (1/3)(3^3) + (3^2) + 8(3) ]

= 2 * [ (216/3) - 36 - 48 - (27/3) + 9 + 24 ]

= 2 * [ 72 - 36 - 48 - 9 + 9 + 24 ]

= 2 * [ 12 ]

= 24 meters

Adding the distances together:

Total distance traveled = |-32| + 24 = 32 + 24 = 56 meters

Therefore, the total distance traveled by the particle is 56 meters.

User Kimichang
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