A car moves along a straight road in such a way that its velocity (in feet per second) at any time t (in seconds) is given by v(x) = 3x sqrt(9-x^2) on an interval of 0,3.
Find the distance traveled by the car in 3 seconds.
v(x) = 3x sqrt(9-x^2) = 3x(9-x^2)^(1/2)
s(x) = -(9-x^2)^(3/2) + c
s(0) = -27 + c
s(3) = 0 + c
distance = (0 + c) - (-27 + c) = 27 ft
Well, isn't this car full of square roots and laughs! Let's figure out how far it traveled in 3 seconds.
To find the distance traveled by the car in 3 seconds, we need to integrate the velocity function over the interval of 0 to 3.
So, let's set up the integral:
∫[0,3] (3x√(9-x^2)) dx
Now, let's evaluate this integral. But hang on, let's not take it too seriously! Remember, we're here to have some fun!
So, we have:
∫[0,3] (3x√(9-x^2)) dx
To integrate this funtastic function, we'll use a little substitution. Let z = 9-x^2. Then, dz = -2x dx.
If we plug in these super fun substitutions, we get:
∫[0,3] (3x√z / -2x) dz
See what we did there? We got rid of that pesky x. Now, let's simplify:
-3/2 ∫[0,3] √z dz
To make things even more exciting, let's change our limits of integration. When x = 0, z = 9. And when x = 3, z = 0.
So, we have:
-3/2 ∫[9,0] √z dz
Now, let's integrate this and let the fun begin:
-3/2 ∫[9,0] z^(1/2) dz
Using the power rule of integration, we get:
-3/2 (2/3) z^(3/2) |[9,0]
Since we're all about having fun with math, let's plug in our limits of integration:
-3/2 (2/3) [(0)^(3/2) - (9)^(3/2)]
Now, let's simplify further:
-3/2 (2/3) [-9^(3/2)]
To make this even more hilarious, let's calculate:
-3/2 (2/3) [-(27 × √9)]
Remember, square root of 9 is just 3. So, we have:
-3/2 (2/3) [-(27 × 3)]
Now, let's multiply and divide like the clowns we are:
-3/2 (2/3) [-81]
Finally, let's compute:
(-3/2) × (2/3) × (-81)
And the result is...
81 feet!
So, in 3 seconds, this wacky car traveled a distance of 81 feet along the straight road. Isn't that just hilarious?
To find the distance traveled by the car in 3 seconds, we need to calculate the definite integral of the velocity function over the interval from 0 to 3.
First, let's rewrite the given velocity function as v(t) = 3t * sqrt(9 - t^2). Since the interval is from 0 to 3, the definite integral can be calculated as follows:
∫[0,3] 3t * sqrt(9 - t^2) dt
To evaluate this integral, we can use the substitution method. Let's substitute u = 9 - t^2 and find du/dt.
u = 9 - t^2
du/dt = -2t
Rearrange the equation to solve for dt:
dt = -du / (2t)
Now, substitute these expressions into the integral:
∫[0,3] 3t * sqrt(9 - t^2) dt
= ∫[u=9,t=0,t=3] 3t * sqrt(u) * ( -du / (2t) )
= - 3/2 ∫[9,0,3] sqrt(u) du
= - 3/2 ∫[0,9] sqrt(u) du
Now, we can evaluate this integral. The integral of sqrt(u) du is (2/3) * (u^(3/2)). Let's substitute back u = 9.
- 3/2 * [ (2/3) * (u^(3/2)) ] [0,9]
= -3/2 * [ (2/3) * (9^(3/2)) - (2/3) * (0^(3/2)) ]
= -3/2 * [ (2/3) * (9 * sqrt(9)) - (2/3) * (0) ]
= -3/2 * [ (2/3) * (9 * 3) - (2/3) * (0) ]
= -3/2 * [ 6 * 9 - 0 ]
= -3/2 * 54
= -81
Since distance cannot be negative, we take the absolute value of the result:
| -81 | = 81
Therefore, the car traveled a distance of 81 feet in 3 seconds.
To find the distance traveled by the car in 3 seconds, we need to calculate the definite integral of the velocity function from 0 to 3. The definite integral represents the area under the curve of the function, which in this case gives us the total distance traveled.
The given velocity function is v(x) = 3x sqrt(9-x^2). To calculate the definite integral, follow these steps:
1. Start by finding the antiderivative of the velocity function. The antiderivative of 3x sqrt(9-x^2) can be found by making a substitution: let u = 9 - x^2. Then differentiate both sides with respect to x to find du/dx = -2x, and solve for dx = -du/(2x). Substitute these values into the original integral:
∫ 3x sqrt(9-x^2) dx = ∫ 3x sqrt(u) (-du/(2x))
Simplifying, we have:
= -3/2 ∫ sqrt(u) du
2. Integrate -3/2 ∫ sqrt(u) du with respect to u. The integral of sqrt(u) is (2/3)u^(3/2). Applying this to the integral:
= -3/2 * (2/3) ∫ u^(3/2) du
Simplifying further, we have:
= -u^(3/2)
3. Substitute back u = 9 - x^2 into the result:
= -(9 - x^2)^(3/2)
4. Evaluate the expression from 0 to 3:
Distance = ∫[0 to 3] -(9 - x^2)^(3/2) dx
= -[(9 - 3^2)^(3/2)] - [0]
= -(9 - 9)^(3/2)
= -(0)^(3/2)
= -0
Therefore, the distance traveled by the car in 3 seconds is 0 feet.