The velocity function is v(t)=t2−3t+2 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [−3,6].
displacement = ??
distance traveled = ??
v(t) = t^2 - 3t + 2
displacement is
s = ∫[-3,6] (t^2 - 3t + 2) dt = 117/2
distance is
∫[-3,6] |t^2 - 3t + 2| dt = 353/6
To find the displacement, we need to find the change in position of the particle. The displacement can be found by integrating the velocity function over the interval [−3, 6].
To find the distance traveled, we need to consider both the forward and backward motion of the particle. So, we need to integrate the absolute value of the velocity function.
First, let's find the displacement:
Displacement = ∫[−3, 6] v(t) dt
Plugging in the given velocity function v(t) = t^2 - 3t + 2, we have:
Displacement = ∫[−3, 6] (t^2 - 3t + 2) dt
Integrating the expression yields:
Displacement = [((t^3)/3) - (3(t^2)/2) + (2t)] between -3 and 6
Substituting the upper and lower limits, we get:
Displacement = [((6^3)/3) - (3(6^2)/2) + (2(6))] - [((-3^3)/3) - (3((-3)^2)/2) + (2(-3))]
Simplifying the expression gives:
Displacement = (72 - 54 + 12) - (-3 + 13 - 6)
Displacement = 30 + 10
Displacement = 40
Therefore, the displacement of the particle during the time interval [−3, 6] is 40 units.
Next, let's find the distance traveled:
Distance travelled = ∫[−3, 6] |v(t)| dt
Plugging in the given velocity function v(t) = t^2 - 3t + 2, we have:
Distance travelled = ∫[−3, 6] |t^2 - 3t + 2| dt
To integrate the absolute value function, we need to consider two cases: when the expression inside the absolute value is positive and when it is negative.
Case 1: t^2 - 3t + 2 ≥ 0
Solving the quadratic inequality, we find the interval of t where the expression inside the absolute value is positive:
(t - 1)(t - 2) ≥ 0
The solution is 1 ≤ t ≤ 2.
Case 2: t^2 - 3t + 2 < 0
Solving the quadratic inequality, we find the interval of t where the expression inside the absolute value is negative:
t < 1 or t > 2.
Now, we can split the integral into two parts based on the cases:
Distance travelled = ∫[1, 2] (t^2 - 3t + 2) dt + ∫[−3, 1] -(t^2 - 3t + 2) dt
Integrating each expression separately, we get:
Distance travelled = [((t^3)/3) - (3(t^2)/2) + (2t)] between 1 and 2 + [(-(t^3)/3) + (3(t^2)/2) - (2t)] between -3 and 1
Substituting the upper and lower limits, we have:
Distance travelled = [((2^3)/3) - (3(2^2)/2) + (2(2))] - [((1^3)/3) - (3(1^2)/2) + (2(1))] + [(-(1^3)/3) + (3(1^2)/2) - (2(1))] - [((-3^3)/3) + (3((-3)^2)/2) - (2(-3))]
Simplifying the expression gives:
Distance travelled = (8/3 - 12 + 4) - (1/3 - 3/2 + 2) + (-1/3 + 3/2 - 2) - (-27/3 + 9/2 - 6)
Distance travelled = (4/3 - 12 + 4) - (1/3 - 3/2 + 2) + (-1/3 + 3/2 - 2) - (-9 + 9/2 - 6)
Distance travelled = (8/3 - 6 + 4) - (1/3 - 3/2 + 2) + (-1/3 + 3/2 - 2) - (-9/2 - 9/2)
Simplifying further, we have:
Distance travelled = (6/3 + 4) - (1/3 - 3/2 + 2) + (-1/3 + 3/2 - 2) - (-(9+9)/2)
Distance travelled = 2 + (1/3 - 3/2 + 2) + (-1/3 + 3/2 - 2) + (-18/2)
Distance travelled = 2 + (6/6 - 9/6 + 12/6) + (-2/6 + 9/6 - 12/6) + (-18/2)
Distance travelled = 2 + (9/6) + (-5/6) + (-9)
Distance travelled = 2 + 1.5 - 0.83 - 9
Distance travelled = -5.33
Therefore, the distance traveled by the particle during the time interval [−3, 6] is approximately 5.33 units.
To find the displacement and distance traveled by the particle, we need to integrate the velocity function over the given time interval.
First, let's find the displacement. Displacement measures the change in position of the particle. It can be found by taking the definite integral of the velocity function from the initial time point to the final time point.
The displacement can be calculated as follows:
1. Integrate the velocity function v(t) with respect to time. In this case, the velocity function is v(t) = t^2 - 3t + 2.
∫(t^2 - 3t + 2) dt = (1/3)t^3 - (3/2)t^2 + 2t + C
2. Evaluate the definite integral by substituting the upper and lower limits of the time interval. In this case, the limits are -3 and 6.
Displacement = [(1/3)(6)^3 - (3/2)(6)^2 + 2(6)] - [(1/3)(-3)^3 - (3/2)(-3)^2 + 2(-3)]
= [(1/3)216 - (3/2)36 + 12] - [(1/3)(-27) - (3/2)9 - 6]
= [72 - 54 + 12] - [-9 - 13 - 6]
= 30 - (-28)
= 30 + 28
= 58
The displacement of the particle during the time interval [-3, 6] is 58 units.
Next, let's find the distance traveled. Distance measures the total path length covered by the particle. It can be calculated by taking the integral of the absolute value of the velocity function over the given time interval.
To find the distance traveled, follow these steps:
1. Take the absolute value of the velocity function |v(t)| = |t^2 - 3t + 2|.
|t^2 - 3t + 2| = t^2 - 3t + 2 if t^2 - 3t + 2 ≥ 0
= -(t^2 - 3t + 2) if t^2 - 3t + 2 < 0
2. Integrate the absolute value of the velocity function with respect to time over the time interval [-3, 6].
∫|t^2 - 3t + 2| dt = ∫(t^2 - 3t + 2) dt for t^2 - 3t + 2 ≥ 0
= ∫-(t^2 - 3t + 2) dt for t^2 - 3t + 2 < 0
3. Evaluate the definite integral by splitting it into two cases based on the sign of t^2 - 3t + 2 within the given time interval.
Case 1: t^2 - 3t + 2 ≥ 0
Integrate t^2 - 3t + 2 from -3 to 6:
∫(t^2 - 3t + 2) dt = [1/3t^3 - 3/2t^2 + 2t] evaluated from -3 to 6
= [(1/3)(6)^3 - (3/2)(6)^2 + 2(6)] - [(1/3)(-3)^3 - (3/2)(-3)^2 + 2(-3)]
= [72 - 54 + 12] - [-9 - 13 - 6]
= 30 - (-28)
= 30 + 28
= 58
Case 2: t^2 - 3t + 2 < 0
Integrate -(t^2 - 3t + 2) from -3 to 6:
∫-(t^2 - 3t + 2) dt = -[(1/3t^3 - 3/2t^2 + 2t)] evaluated from -3 to 6
= -[(1/3)(6)^3 - (3/2)(6)^2 + 2(6)] + [(1/3)(-3)^3 - (3/2)(-3)^2 + 2(-3)]
= -[72 - 54 + 12] + [-9 - 13 - 6]
= -30 + (-28)
= -30 - 28
= -58
The distance traveled by the particle during the time interval [-3, 6] is the absolute value of the sum of the two cases:
Distance traveled = |Case 1| + |Case 2|
= |58| + |-58|
= 58 + 58
= 116
So, the distance traveled by the particle during the time interval [-3, 6] is 116 units.