A small 7.10kg rocket burns fuel that exerts a time-varying upward force on the rocket. This force obeys the equation F = A + Bt^2. Measurements show that at t = 0, the force is 121.0N, and at the end of the first 2.10s , it is 176.0N.
Part A
Find the net force on this rocket the instant after the fuel ignites.
Part B
Find the acceleration of this rocket the instant after the fuel ignites.
Part C
Find the net force on this rocket 4.10s after fuel ignition.
Part D
Find the acceleration of this rocket 4.10s after fuel ignition.
I found out that
A = 121.0 N
B = 12.5 N/s^2
Net Force = 51.42 N
Acceleration = 7.24 m/s^2
Part C and Part D is what I am having trouble with.
By the way, the A = 121.0 and B = 12.5 refers to the equation F=A+Bt^2, not Part A and Part B.
A, B, are correct.
a) and b) are also correct.
For part (c), the question is asking for the force, F(t) = A+Bt², using the values of A and B that you have found earlier, and t=4.1 s. F(4.1) should be over 300 m/s/s.
Once you have calculated the force, you can proceed to calculate the acceleration as in part b).
For Part C, I calculated it to be 331.125N, but am still getting a wrong answer. I know you said it was over 300, is that what you got too?
I have the same number as you have, but based on your values of A and B, which I found minor discrepancies. If it is a computer answer, you may want to recheck from there on.
Please show working
Part C:
To find the net force on the rocket 4.10s after fuel ignition, we can substitute the given values into the equation F = A + Bt^2.
Using t = 4.10s, A = 121.0N, and B = 12.5 N/s^2, we get:
F = A + Bt^2
F = 121.0N + (12.5 N/s^2)(4.10s)^2
F = 121.0N + (12.5 N/s^2)(16.81s^2)
F = 121.0N + 210.125N
F = 331.125N
So, the net force on the rocket 4.10s after fuel ignition is approximately 331.125N.
Part D:
To find the acceleration of the rocket 4.10s after fuel ignition, we can use Newton's second law of motion, F = ma, where F is the net force and a is the acceleration.
Using F = 331.125N (net force from Part C) and the mass of the rocket is given as 7.10kg, we get:
F = ma
331.125N = (7.10kg)a
a = 331.125N / 7.10kg
a ≈ 46.57 m/s^2
So, the acceleration of the rocket 4.10s after fuel ignition is approximately 46.57 m/s^2.
To find the net force on the rocket 4.10s after fuel ignition and the acceleration of the rocket at that time, we can use the given equation for force: F = A + Bt^2, where A = 121.0 N and B = 12.5 N/s^2.
Part C:
To find the net force on the rocket 4.10s after fuel ignition, we need to substitute the value of t into the equation and solve for F.
At t = 4.10s,
F = A + Bt^2 = 121.0 N + 12.5 N/s^2 * (4.10s)^2
First, we need to square the value of 4.10s:
t^2 = 4.10s * 4.10s = 16.81s^2
Now, we substitute this value into the equation:
F = 121.0 N + 12.5 N/s^2 * 16.81s^2
Simplifying this expression, we get:
F = 121.0 N + 12.5 N/s^2 * 16.81s^2
F ≈ 335.762 N
So, the net force on the rocket 4.10s after fuel ignition is approximately 335.762 N.
Part D:
To find the acceleration of the rocket 4.10s after fuel ignition, we can use Newton's second law of motion, which states that F = ma, where F is the net force and a is the acceleration.
Since we know the net force F (as calculated in Part C) and the mass of the rocket (7.10 kg), we can rearrange the equation to solve for acceleration:
a = F/m
Plugging in the values:
a = 335.762 N / 7.10 kg
Evaluating this expression, we get:
a ≈ 47.31 m/s^2
So, the acceleration of the rocket 4.10s after fuel ignition is approximately 47.31 m/s^2.