B. Tornado destruction
It is based on the amount of damage
<span>(a) 12.02 m/s
(b) 52.2 meters
This problem is an example of integral calculus. You've been given an acceleration vector which is usually known as the 2nd derivative. From that you need to calculate the velocity function (1st derivative) and position (actual function) by successively calculating the anti-derivative. So:
A(t) = 6.30 - 2.20t
V(t) = 6.30t - 1.10t^2 + C
We now have a velocity function, but need to determine C. Since we've been given the velocity at t = 0, that's fairly trivial.
V(t) = 6.30t - 1.10t^2 + C
3 = 6.30*0 - 1.10*0^2 + C
3 = 0 + 0 + C
3 = C
So the entire velocity function is:
V(t) = 6.30t - 1.10t^2 + 3
V(t) = -1.10t^2 + 6.30t + 3
Now for the location function which is the anti-derivative of the velocity function.
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
Now we need to calculate C. And once again, we've been given the location for t = 0, so
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + C
7.3 = -0.366666667*0^3 + 3.15*0^2 + 3*0 + C
7.3 = 0 + 0 + 0 + C
7.3 = C
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
Now that we have the functions, they are:
A(t) = 6.30 - 2.20t
V(t) = -1.10t^2 + 6.30t + 3
L(t) = -0.366666667t^3 + 3.15t^2 + 3t + 7.3
let's answer the questions.
(a) What is the maximum speed achieved by the cyclist?
This can only happen at those points that meet either of the following criteria.
1. The derivative is undefined for the point.
2. The value of the derivative is 0 for the point.
As it turns out, the 1st derivative of the velocity function is the acceleration function which we have. So
A(t) = 6.30 - 2.20t
0 = 6.30 - 2.20t
2.20t = 6.30
t = 2.863636364
So one of V(0), V(2.863636364), or V(6) will be the maximum value. Therefore:
V(0) = 3
V(2.863636364) = 12.0204545454545
V(6) = 1.2
So the maximum speed achieved is 12.02 m/s
(b) Total distance traveled?
L(0) = 7.3
L(6) = 59.5
Distance traveled = 59.5 m - 7.3 m = 52.2 meters</span>
Answer: something must drop it over
Explanation:
Answer:
a) 0.049 m
b) Yes, increase
Explanation:
Draw a free body diagram.
In the y direction, there are three forces acting on the feeder. Two vertical components of the tension forces in each rope pulling up, and weight force pulling down.
Apply Newton's second law to the feeder in the y direction.
∑F = ma
2Ty − mg = 0
Ty = mg/2
Let's say the distance the rope sags is d. The trees are 4m apart, so the feeder is 2m horizontally from either tree. Using Pythagorean theorem, we can find the length of the rope on either side:
L² = 2² + d²
L = √(4 + d²)
Using similar triangles, we can write a proportion using the forces and distances.
Ty / T = d / L
Substitute:
(mg/2) / T = d / √(4 + d²)
Solve for d:
Td = mg/2 √(4 + d²)
T² d² = (mg/2)² (4 + d²)
T² d² = (mg)² + (mg/2)² d²
(T² − (mg/2)²) d² = (mg)²
d² = (mg)² / (T² − (mg/2)²)
d = mg / √(T² − (mg/2)²)
Given m = 2.4 kg and T = 480 N:
d = (2.4) (9.8) / √(480² − (2.4×9.8/2)²)
d = 0.049 m
b) If a bird lands on a feeder, this will increase the tension in the rope to support the bird's weight.
