The velocity function (in meters per second) for a particle moving along a line is given by v(t)=t3–4t2. Find the displacement and the distance traveled by the particle during the time interval [-1,6]. (Hint: Draw a graph of the velocity function.) Your answers require that you enter the correct units.
displacement found to be 34.4166 m
help finding distance please!!!!
To find the distance traveled by the particle during the time interval [-1, 6], we need to integrate the absolute value of the velocity function over that interval. This will give us the total distance traveled, regardless of the direction.
First, let's visualize the velocity function by graphing it. The velocity function is v(t) = t^3 - 4t^2.
To graph the function, we can plot a few points and connect them to get an idea of the shape. Let's calculate some values of v(t) for different values of t:
When t = -1, v(-1) = (-1)^3 - 4(-1)^2 = -1 + 4 = 3
When t = 0, v(0) = 0^3 - 4(0)^2 = 0
When t = 1, v(1) = 1^3 - 4(1)^2 = 1 - 4 = -3
When t = 2, v(2) = 2^3 - 4(2)^2 = 8 - 16 = -8
When t = 3, v(3) = 3^3 - 4(3)^2 = 27 - 36 = -9
When t = 4, v(4) = 4^3 - 4(4)^2 = 64 - 64 = 0
Plotting these points on a graph, we can connect them to get a rough sketch of the velocity function. The graph should resemble a downward sloping curve that starts above the x-axis, dips below it, and then returns to the x-axis.
Now, to find the distance traveled, we integrate the absolute value of the velocity function over the interval [-1, 6]. This will give us the area under the curve, representing the total distance traveled.
Let's set up the definite integral:
distance = ∫(|v(t)|)dt from -1 to 6
Since the velocity function changes signs at t = 0 and t = 4, we can break the interval into three parts: [-1, 0], [0, 4], and [4, 6].
For the interval [-1, 0]:
distance1 = ∫(|v(t)|)dt from -1 to 0
distance1 = ∫(3)dt from -1 to 0
distance1 = 3t from -1 to 0
distance1 = 3(0) - 3(-1)
distance1 = 3
For the interval [0, 4]:
distance2 = ∫(|v(t)|)dt from 0 to 4
distance2 = ∫(t^3 - 4t^2)dt from 0 to 4
To evaluate this integral, we need to find the antiderivative of (t^3 - 4t^2):
∫(t^3 - 4t^2)dt = (1/4)t^4 - (4/3)t^3 + C
Applying the antiderivative, we have:
distance2 = [(1/4)t^4 - (4/3)t^3] from 0 to 4
distance2 = [(1/4)(4)^4 - (4/3)(4)^3] - [(1/4)(0)^4 - (4/3)(0)^3]
distance2 = (1/4)(256) - (4/3)(64)
distance2 = 64 - 85.333
distance2 = -21.333
For the interval [4, 6]:
distance3 = ∫(|v(t)|)dt from 4 to 6
distance3 = ∫(t^3 - 4t^2)dt from 4 to 6
Using the antiderivative, we have:
distance3 = [(1/4)t^4 - (4/3)t^3] from 4 to 6
distance3 = [(1/4)(6)^4 - (4/3)(6)^3] - [(1/4)(4)^4 - (4/3)(4)^3]
distance3 = (1/4)(1296) - (4/3)(216) - (1/4)(256) + (4/3)(64)
distance3 = 324 - 288 - 64 + 21.333
distance3 = -6.667
Now, we add up the distances for each interval to find the total distance traveled:
distance = distance1 + distance2 + distance3
distance = 3 + (-21.333) + (-6.667)
distance = -25
The distance traveled by the particle during the time interval [-1, 6] is 25 meters. Note that the distance is always positive, regardless of the negative values obtained during the calculations.
Therefore, the distance traveled by the particle is 25 meters.