A particle moves in a straight line with velocity t^-2 - 1/9 ft/s. Find the total displacement and total distance traveled over the time interval [1,4].
I found out that the total displacement is .4116
But I cannot find the total distance traveled.
Let
v(t)=velocity function = t^-2 - 1/9
s(t)=displacement function. = -t/9-1/t
and
displacement = s(4)-s(1)= -25/36 - (-10/9) = 5/12 ft.
If s(t) is monotonically increasing or decreasing, then the displacement equals the distance travelled.
However, we note that v(3)=0, after which time the velocity reverses in direction.
So the distance travelled equals
s(4)-s(3) - [s(3)-s(1)]
=-25/36 -(-2/3) - [(-2/3)-(-10/9)]
=-17/36
Ignore the sign, since distance is a scalar.
So distance = 17/36 ft.
Well, finding the total distance can be a bit tricky when dealing with varying velocities. But fear not, I, Clown Bot, am here to help!
To find the total distance traveled, you'll need to integrate the absolute value of the velocity function over the given time interval. In this case, the velocity function is t^-2 - 1/9 ft/s, with t ranging from 1 to 4.
Now, let's integrate the absolute value of the function. But be warned, integration can be a bit "integral" to understand!
∫|t^-2 - 1/9| dt, from 1 to 4.
Now, let's take a closer look at the function inside the absolute value. Notice that for values of t greater than 1, t^-2 is always positive, and 1/9 is positive as well. So the absolute value isn't required in this case.
∫(t^-2 - 1/9) dt, from 1 to 4.
Before we proceed, let's compute the antiderivative of (t^-2 - 1/9):
∫(t^-2 - 1/9) dt = -t^-1/3 - t/9.
Now, let's evaluate this expression between 1 and 4:
[-(4)^-1/3 - (4)/9] - [-(1)^-1/3 - (1)/9].
Now, time for some calculations!
[(-1/∛4) - (4/9)] - [(-1/∛1) - (1/9)].
Simplifying:
[(-1/∛4) - (4/9)] - [(-1/∛1) - (1/9)].
(-(1/∛4) - (4/9)) - (-(1/∛1) - (1/9)).
Now, let's simplify this even further:
(-(1/∛4) - (4/9)) - (-(1) - (1/9)).
(-(1/∛4) - (4/9)) - (-(1)-(1/9)).
Turning the subtraction into addition:
(-(1/∛4) - (4/9)) + (1+(1/9)).
Now, let's add the terms inside each grouping:
(-(1/∛4) - (4/9)) + (1+(1/9)).
Combining like terms:
-(1/∛4) -(4/9) + 1 + (1/9).
Finding a common denominator for addition:
-(1/∛4) - (4/9) + (9/9) + (1/9).
Adding:
-(1/∛4) - (4/9) + (9+1+1)/9.
Combining like terms:
-(1/∛4) - (4/9) + 11/9.
Finally, simplifying:
-(1/∛4) - (4/9) + 11/9.
Using a common denominator:
-9/(9∛4) - 4∛4/9 + 11/9.
Laughing through the math:
-9/(9∛4) - 4∛4/9 + 11/9.
Simplifying further:
-1/∛4 - 4∛4/9 + 11/9.
That's your final answer for the total distance traveled! Now, I hope this "integral" solution brought a smile to your face. Keep up the good work!
To find the total distance traveled, we need to calculate the absolute value of the velocity function and integrate it over the given time interval.
The velocity function is given as v(t) = t^-2 - 1/9 ft/s.
To find the total distance traveled, we need to calculate the integral of the absolute value of the velocity function over the time interval [1,4].
∫|v(t)| dt = ∫|t^-2 - 1/9| dt
Since the velocity function is negative for certain values of t, we need to break up the integral into two parts, one from 1 to the point where the velocity function becomes 0, and another from that point to 4.
From 1 to the point where the velocity function becomes 0:
∫(1 to √(9/8)) (t^-2 - 1/9) dt = ∫(1 to √(9/8)) t^-2 dt - ∫(1 to √(9/8)) (1/9) dt
From the point where the velocity function becomes 0 to 4:
∫(√(9/8) to 4) -(t^-2 - 1/9) dt
Performing the integration and taking the absolute values, we get:
∫(1 to √(9/8)) t^-2 dt - ∫(1 to √(9/8)) (1/9) dt + ∫(√(9/8) to 4) (t^-2 - 1/9) dt
Simplifying further:
[-t^-1] from 1 to √(9/8) - [(1/9)t] from 1 to √(9/8) - [-t^-1 + (1/9)t] from √(9/8) to 4
Evaluating each part:
[-(√(8/9) - 1)] - [(1/9)√(8/9) - (1/9)] - [-4^-1 + (1/9)(4)] + [(√(8/9))^(-1) - (1/9)√(8/9)]
Simplifying further:
[-(√(8/9) - 1)] - [(1/9)√(8/9) - (1/9)] + [3/4 + 4/9] + [(√(8/9))^(-1) - (1/9)√(8/9)]
Now, substitute the values and calculate:
[-(√(8/9) - 1)] - [(1/9)√(8/9) - (1/9)] + [3/4 + 4/9] + [(√(8/9))^(-1) - (1/9)√(8/9)]
≈ -0.364 + 0.101 + 1.083 + 0.765
≈ 1.585 ft
Therefore, the total distance traveled over the time interval [1,4] is approximately 1.585 ft.
To find the total distance traveled over the time interval [1,4], you need to integrate the absolute value of the velocity function over that time interval.
Let's start by finding the absolute value of the velocity function for the given time interval. The absolute value function removes the negative sign from any negative values. In this case, the absolute value of t^-2 - (1/9) is |t^-2 - (1/9)|.
Next, integrate the absolute value of the velocity function over the time interval [1,4]. The integral of a function gives you the area under the curve of that function. The definite integral represents the area between the function and the x-axis over a specific interval.
Integrate |t^-2 - (1/9)| with respect to t from 1 to 4:
∫(1 to 4) |t^(-2) - (1/9)| dt
This integral can be split into two separate integrals, each corresponding to different intervals where the function is positive and negative:
∫(1 to 4) (t^(-2) - (1/9)) dt (for t from 1 to 3)
∫(1 to 4) ((-t^(-2)) + (1/9)) dt (for t from 3 to 4)
Simplify and evaluate each integral separately:
First integral:
∫(1 to 3) (t^(-2) - (1/9)) dt
= [ -t^(-1) - (t/9) ] (from 1 to 3)
= [-3^(-1) - (3/9)] - [-1^(-1) - (1/9)]
= [-1/3 - 1/3] - [-1 - 1/9]
= -2/3 + 1/9
Second integral:
∫(3 to 4) ((-t^(-2)) + (1/9)) dt
= [ (t^(-1)) + (t/9) ] (from 3 to 4)
= [4^(-1) + (4/9)] - [3^(-1) + (3/9)]
= [1/4 + 4/9] - [1/3 + 1/9]
= 9/36 + 16/36 - 12/36 - 4/36
= 9/36
Now add the results of both integrals to find the total distance traveled over the interval [1,4]:
Total distance = (-2/3 + 1/9) + (9/36)
= -2/3 + 1/9 + 9/36
= -8/12 + 4/12 + 9/36
= 5/12 + 9/36
= 15/36
= 5/12 ft
Therefore, the total distance traveled over the time interval [1,4] is 5/12 ft.