Prof gave us this question to practice but I do not know how to solve it. If you know please provide step by step with the answer so I can understand it. Thank you very much :)

Suppose a rocket is launched from the ground with 10 seconds worth of fuel. The rocket has an upward acceleration of 8 m/s^2 while it still has fuel but after the fuel runs out, it has an acceleration of −9.8 m/s^2.

(a) Find functions describing the velocity and position of the rocket while it still has fuel.
(b) Find the velocity and height of the rocket at the moment it runs out of fuel.
(c) Find functions describing the velocity and position of the rocket after it has run
out of fuel.
(d) Find the time when the maximum height is reached by the rocket. What is the maximum height (round to one decimal place)?

under power,

a(t) = 8
v(t) = 8t
h(t) = 4t^2
So,
v(10) = 80
h(10) = 400

Now, in ballistic path, only gravity affects the motion, so for t>10,

h(t) = 400 + 80(t-10) - 9.8(t-10)^2
= -1380 + 276t - 9.8t^2

Now you can find the vertex, the roots, etc.

hahahahahahah UW Math 127 assignment

lolimsodead

Bonjour....

To solve this problem, we can use the principles of kinematics and integrate the given information. I'll walk you through the steps to solve each part of the question.

(a) Find functions describing the velocity and position of the rocket while it still has fuel.

To find the velocity function, we integrate the given acceleration function while the rocket still has fuel:
∫(8) dt = 8t + C1

Where t represents time and C1 is the constant of integration. Since the rocket starts from rest (at t=0), we can set the initial velocity (v0) to 0.

v(t) = 8t + C1

To find the position function, we integrate the velocity function:
∫(8t + C1) dt = 4t^2 + C1t + C2

Where C2 is the constant of integration. Since the initial position (x0) is also 0, we can set C2 to 0.

x(t) = 4t^2 + C1t

(b) Find the velocity and height of the rocket at the moment it runs out of fuel.

The rocket runs out of fuel after 10 seconds, so we'll substitute t = 10 into the velocity and position functions.

v(10) = 8(10) + C1 = 80 + C1

Since we don't know the exact fuel consumption, we can consider C1 as a constant specific to this problem.

For height, we find the position function at t = 10.

x(10) = 4(10^2) + C1(10) = 400 + 10C1

(c) Find functions describing the velocity and position of the rocket after it has run out of fuel.

After the fuel runs out, the rocket experiences a different acceleration of -9.8 m/s^2. We can now integrate this new acceleration to find the velocity and position functions.

Integrating -9.8 with respect to time gives us:
∫(-9.8) dt = -9.8t + C3

Where C3 is the constant of integration. Since the velocity of the rocket at t = 10 is v(10) = 80 + C1, we can substitute this value into the new velocity function:

v(t) = -9.8t + (80 + C1)

Integrating the velocity function, we get the position function:
∫(-9.8t + (80 + C1)) dt = -4.9t^2 + (80t + C1t + C4)

Where C4 is the constant of integration. Since the new starting position will be the height of the rocket when it runs out of fuel, we substitute x(10) = 400 + 10C1 into the position function:

x(t) = -4.9t^2 + (80t + C1t) + (400 + 10C1)

(d) Find the time when the maximum height is reached by the rocket. What is the maximum height (round to one decimal place)?

To find the time when the rocket reaches its maximum height, we need to find when the velocity becomes zero. Set v(t) = 0 and solve for t:

-9.8t + (80 + C1) = 0
-9.8t = -80 - C1
t = (80 + C1)/9.8

Substituting this value of t into the position function will give us the maximum height.

x(t) = -4.9t^2 + (80t + C1t) + (400 + 10C1)

For the maximum height, substitute t with (80 + C1)/9.8:

x_max = -4.9((80 + C1)/9.8)^2 + (80((80 + C1)/9.8) + C1((80 + C1)/9.8)) + (400 + 10C1)

Simplify this expression to find the maximum height.

I hope this step-by-step explanation helps you solve the problem. Let me know if you have any further questions!