The velocity v(t)= -t^2 + 6t - 8, find the displacement and distance traveled during the time interval [-1,5]?
Well, since you're dealing with velocity, the displacement is the integral of velocity over the given interval. So let's integrate the velocity function and find the antiderivative.
v(t) = -t^2 + 6t - 8
To integrate -t^2, we add 1 to the exponent and divide by the new exponent (-t^2 -> -t^3/3).
Similarly, integrating 6t gives us 3t^2, and integrating -8 gives us -8t.
So the antiderivative of v(t) is:
s(t) = -t^3/3 + 3t^2 - 8t + C
Now, let's find the displacement by evaluating the antiderivative at the endpoints [-1, 5].
s(5) - s(-1) = [-(5)^3/3 + 3(5)^2 - 8(5) + C] - [-(1)^3/3 + 3(1)^2 - 8(1) + C]
Simplifying this expression, we have:
[-(125/3) + 75 - 40 + C] - [-1/3 + 3 - 8 + C]
Now, the constant C cancels out in the subtraction.
Simplifying further, we get:
[-(125/3) + 75 - 40] - [-1/3 + 3 - 8]
Which simplifies to:
[-14/3] - [-14]
Finally, we get:
-14/3 + 14/1
This equals:
(14 - 42)/3
Which simplifies to:
-28/3
So, the displacement over the interval [-1, 5] is -28/3.
Now, to find the distance traveled, we can take the absolute value of the displacement. Since distance can't be negative, we end up with a positive value of 28/3.
So, during the time interval [-1, 5], the displacement is -28/3 units and the distance traveled is 28/3 units.
To find the displacement and distance traveled during the time interval [-1,5], we first need to find the position function by integrating the velocity function.
The position function can be found by integrating the velocity function v(t) = -t^2 + 6t - 8 with respect to time:
s(t) = ∫ v(t) dt
Integrating, we get:
s(t) = - (1/3) * t^3 + 3t^2 - 8t + C
Now, to find the constant C, we use the given time interval [-1, 5].
Since we are only interested in the displacement and distance traveled during this interval, we can ignore the constant C.
Using the position function, we can find the displacement by evaluating s(t) at the endpoints of the interval:
Displacement = s(5) - s(-1)
s(5) = - (1/3) * 5^3 + 3 * 5^2 - 8 * 5
s(5) = -125/3 + 75 - 40 = -125/3 + 75 - 120/3 = - 170/3
s(-1) = - (1/3) * (-1)^3 + 3 * (-1)^2 - 8 * (-1)
s(-1) = -1/3 + 3 + 8 = -1/3 + 9/3 + 24/3 = 32/3
Displacement = (- 170/3) - (32/3) = - 202/3
To find the distance traveled, we need to take the absolute value of the displacement:
Distance = |Displacement| = | - 202/3| = 202/3
Therefore, the displacement during the time interval [-1,5] is -202/3 and the distance traveled during this interval is 202/3.
To find the displacement and distance traveled, we need to integrate the given velocity function over the given time interval.
First, let's find the displacement. Displacement is represented by the definite integral of velocity over the given time interval:
Displacement = ∫[a,b] v(t) dt
In this case, the time interval is [-1,5], so the equation becomes:
Displacement = ∫[-1,5] (-t^2 + 6t - 8) dt
To evaluate this integral, we need to integrate each term separately:
Displacement = ∫[-1,5] -t^2 dt + ∫[-1,5] 6t dt - ∫[-1,5] 8 dt
Integrating each term:
Displacement = [(-1/3)t^3] from -1 to 5 + [3t^2] from -1 to 5 - [8t] from -1 to 5
Evaluating the definite integral at the upper and lower limits:
Displacement = [(-1/3)(5)^3 - (-1/3)(-1)^3] + [3(5)^2 - 3(-1)^2] - [8(5) - 8(-1)]
Simplifying:
Displacement = [-125/3 + 1/3] + [75 + 3] - [40 + 8]
Displacement = [-124/3] + [78] - [48]
Displacement = -124/3 + 78 - 48
Therefore, the displacement during the time interval [-1,5] is -124/3 + 78 - 48.
To find the distance traveled, we need to consider the absolute value of the velocity function and integrate it over the given time interval:
Distance = ∫[a,b] |v(t)| dt
In this case, the velocity function (-t^2 + 6t - 8) is negative between the roots of the equation (-1 and 5). Therefore, we need to split the integral into two parts:
Distance = ∫[-1,0] -(t^2 - 6t + 8) dt + ∫[0,5] (t^2 - 6t + 8) dt
Integrating each part separately:
Distance = [-t^3/3 + 3t^2 - 8t] from -1 to 0 + [t^3/3 - 3t^2 + 8t] from 0 to 5
Evaluating the definite integral at the upper and lower limits:
Distance = [-0^3/3 + 3(0)^2 - 8(0)] - [-(-1)^3/3 + 3(-1)^2 - 8(-1)] + [(5)^3/3 - 3(5)^2 + 8(5)] - [0^3/3 - 3(0)^2 + 8(0)]
Simplifying:
Distance = [0] - [1/3 + 3 + 8] + [125/3 - 3(25) + 40] - [0]
Distance = -12/3 - [1/3 + 3 + 8] + [125/3 - 75 + 40]
Distance = -4 - [12/3] + [15/3]
Distance = -4 - 4 + 5
Therefore, the distance traveled during the time interval [-1,5] is -4 - 4 + 5.
For the total displacement during that interval, Integrate the velocity function vs time, for -1 < t < 5 s.
Note that v starts out negative at
t = -1, and then becomes 0 at t = 2 and 4 s. Those are "turnaround times". To get the distance travelled (in either direction), you have to add the absolute value of the displacements during these three intervals:
-1 to +2 s
2 to 4 s
4 to 5 s.
You will have to integrate v dt during each of those intervals and add the absolute values.