For an object whose velocity in ft/sec is given by v(t) = -t^2 + 6, what is its displacement, in feet, on the interval t = 0 to t = 6 secs?
-36
36
0
55.596
For an object whose velocity in ft/sec is given by v(t) = -2t^2 + 4, what is its distance travelled, in feet, on the interval t = 0 to t = 2 secs?
4.876
2.667
9.752
2.438
Please help with these 2 questions.
The first one is -36, the second one is NOT 2.667. (FLVS)
d=int v(t)dt=int (- t^2+6)=
= -1/3 t^3+6t over limits
= -1/3(6^3)+36
= -1/3(216)+36=-72+36
The base of a solid in the xy-plane is the circle x2 + y2 = 16. Cross sections of the solid perpendicular to the y-axis are squares. What is the volume, in cubic units, of the solid?
Can you please help me out on this one Bobby?
Also, for the second question I got 2.667 which I think is right.
To find the displacement of an object on a given interval, we need to integrate its velocity function with respect to time over that interval.
For the first question:
The velocity function is given as v(t) = -t^2 + 6. To find the displacement, we need to integrate this function over the interval t = 0 to t = 6 seconds.
∫(-t^2 + 6) dt
To integrate, apply the power rule:
∫(-t^2 + 6) dt = -((1/3)t^3) + 6t
Now, evaluate the definite integral from t = 0 to t = 6 seconds:
Displacement = (-((1/3) * 6^3) + 6 * 6) - (-((1/3) * 0^3) + 6 * 0)
Displacement = (-72 + 36) - (0 + 0)
Displacement = -36 + 0
Displacement = -36 feet
Therefore, the displacement of the object on the interval t = 0 to t = 6 secs is -36 feet.
For the second question:
The velocity function is given as v(t) = -2t^2 + 4. To find the distance traveled, we need to compute the definite integral of the absolute value of the velocity function over the interval t = 0 to t = 2 seconds.
∫|(-2t^2 + 4)| dt
Split the integral into two parts based on the absolute value:
∫(-2t^2 + 4) dt if -2t^2 + 4 ≥ 0
∫(2t^2 - 4) dt if -2t^2 + 4 < 0
The integral ∫(2t^2 - 4) dt represents the distance traveled while the object is moving backwards. Since the question asks for the distance traveled (which is always non-negative), we only need to evaluate the first integral.
Apply the power rule to integrate -2t^2 + 4:
∫(-2t^2 + 4) dt = -((2/3)t^3) + 4t
Now, evaluate the definite integral from t = 0 to t = 2 seconds:
Distance = (-((2/3) * 2^3) + 4 * 2) - (-((2/3) * 0^3) + 4 * 0)
Distance = (-((2/3) * 8) + 8) - (0 + 0)
Distance = (-16/3 + 8) - 0
Distance = -16/3 + 24/3
Distance = 8/3 feet
Therefore, the distance traveled by the object on the interval t = 0 to t = 2 secs is approximately 2.667 feet.
The answers to the questions are:
1. The displacement of the object on the interval t = 0 to t = 6 secs is -36 feet.
2. The distance traveled by the object on the interval t = 0 to t = 2 secs is approximately 2.667 feet.