The velocity function (in meters per second) for a particle moving along a line is given by v(t)=t3−5t2. Find the displacement and the distance traveled by the particle during the time interval [-1,6].
Distance traveled = ??
I got 2089/12m as answer but it was incorrect please help me!
s(t) = 1/4 t^3 - 5/3 t^3 + c
The displacement is s(6)-s(-1)
The distance traveled is the arc length. So, what steps did you follow?
if v(t) = t^3 - 5t^2
s(t) = (1/4)t^4 - (5/3)t^3 + c
s(6) = (1/4)6^4 - (5/3)6^3 = -36 + c
s(-1) = (1/4)(-1)^4 - (5/3)(-1)^3 = 23/12 + c
distance traveled = 23/12+c - (-36+c) = 4550/12
To find the distance traveled by the particle during the time interval [-1, 6], we need to calculate the definite integral of the absolute value of the velocity function.
Given: v(t) = t^3 - 5t^2
To find the displacement, we need to evaluate the definite integral of v(t) from -1 to 6:
∫[from -1 to 6] (t^3 - 5t^2) dt
To evaluate this integral, we need to find the antiderivative of v(t) first.
The antiderivative of t^3 is (1/4) t^4.
The antiderivative of -5t^2 is (-5/3) t^3.
Now, let's evaluate the definite integral:
∫[from -1 to 6] (t^3 - 5t^2) dt = [(1/4) t^4 - (5/3) t^3] evaluated from -1 to 6
Plugging in the upper and lower limits:
[(1/4) (6^4) - (5/3) (6^3)] - [(1/4) (-1^4) - (5/3) (-1^3)]
Simplifying the expression:
[(1/4) (1296) - (5/3) (216)] - [(1/4) - (5/3)]
= [324 - (360)] - [(1/4) + (5/3)]
= -36 - (47/12)
= - 36 - (47/12)
= -459/12
Therefore, the displacement of the particle during the time interval [-1, 6] is -459/12 meters.
To find the distance traveled, we need to consider the absolute value of the velocity function. Therefore, for any values of t where v(t) is negative, we need to multiply it by -1.
Looking at the expression v(t) = t^3 - 5t^2, we can see that v(t) is negative for 0 ≤ t ≤ 5.
So, the distance traveled is given by the following definite integral:
∫[from -1 to 0] (-(t^3 - 5t^2)) dt + ∫[from 0 to 5] (t^3 - 5t^2) dt + ∫[from 5 to 6] (-(t^3 - 5t^2)) dt
Evaluating this integral, we can follow the same steps as before to find the distance traveled by the particle during the time interval [-1, 6].
To find the distance traveled by the particle during the time interval [-1,6], we need to calculate the total distance covered, which can be done by integrating the absolute value of the velocity function over that interval.
To start, let's calculate the displacement first. Displacement is the change in position from the initial to the final position. It is given by integrating the velocity function over the given interval.
∫(v(t)) dt = ∫(t^3 - 5t^2) dt
To find an antiderivative (integral) of this function, use the power rule of integration:
∫(t^3 - 5t^2) dt = (1/4)t^4 - (5/3)t^3 + C
From -1 to 6:
Displacement = [(1/4)(6^4) - (5/3)(6^3)] - [(1/4)(-1^4) - (5/3)(-1^3)]
Displacement = [(1/4)(1296) - (5/3)(216)] - [(1/4)(1) - (5/3)(-1)]
Displacement = [324 - 360] - [1/4 + 5/3]
Displacement = [-36] - [3/12 + 20/12]
Displacement = -36 + (-23/12)
Displacement = -432/12 - 23/12
Displacement = -455/12 meters
Now, let's calculate the distance traveled. Distance is always positive and accounts for any back-and-forth movement.
To find the distance traveled, we take the integral of the absolute value of the velocity function over the interval [-1, 6]:
∫|v(t)| dt = ∫|t^3 - 5t^2| dt
To integrate the absolute value, we need to split the integral into two cases, depending on the sign of the integrand. So, we calculate the integral of the positive portion and the integral of the negative portion separately.
Case 1: t^3 - 5t^2 ≥ 0 (t^3 - 5t^2 is positive)
∫(t^3 - 5t^2) dt = (1/4)t^4 - (5/3)t^3 + C1
Case 2: t^3 - 5t^2 < 0 (t^3 - 5t^2 is negative)
∫-(t^3 - 5t^2) dt = -(1/4)t^4 + (5/3)t^3 + C2
Now, we evaluate the integral in each case over the interval [-1, 6]:
∫|v(t)| dt = ∫(t^3 - 5t^2) dt (for t ∈ [-1, 6])
= ∫(t^3 - 5t^2) dt (for t ∈ [-1, 0]) + ∫(t^3 - 5t^2) dt (for t ∈ [0, 6])
= [(1/4)(0^4) - (5/3)(0^3)] - [(1/4)((-1)^4) - (5/3)((-1)^3)] + [(1/4)(6^4) - (5/3)(6^3)] - [(1/4)(0^4) - (5/3)(0^3)]
= [0 - 0] - [(1/4)(1) - (5/3)(-1)] + [(1/4)(1296) - (5/3)(216)] - [0 - 0]
= 0 + (23/12) + (324 - 360) + 0
= (23/12) + (-36)
= -23/12 - 36/1
= -23/12 - 432/12
= -455/12 meters
Therefore, the correct answer for the distance traveled is -455/12 meters, which means that the particle traveled a total distance of 455/12 meters during the time interval [-1, 6].