A small 6.90-kg rocket burns fuel that exerts a time-varying upward force on the rocket. This force obeys the equation F=A+Bt2. Measurements show that at t=0, the force is 119.0 N , and at the end of the first 2.40 s , it is 193.0 N .
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 3.90s after fuel ignition.
Part D
Find the acceleration of this rocket 3.90s after fuel ignition.
I found out that
A = 119.0 N
B = 12.8 N/s^2
Net Force = 51.4 N
Acceleration = 7.45 m/s^2
Part C and Part D is what I am having trouble with.
same problem cant solve part C and D
To find the net force on the rocket 3.90s after fuel ignition, we can use the equation F = A + Bt^2, where A is the initial force and B is the rate of change of force with respect to time.
Given:
Initial force, F(0) = 119.0 N
Rate of change of force, B = 12.8 N/s^2
To find the net force at 3.90s, we substitute t = 3.90s into the equation:
F = A + Bt^2
Since A is given as 119.0 N and B is given as 12.8 N/s^2, we have:
F = 119.0 N + 12.8 N/s^2 * (3.90s)^2
Calculating the expression inside the brackets first:
(3.90s)^2 = 15.21s^2
Substituting this back into the equation:
F = 119.0 N + 12.8 N/s^2 * 15.21s^2
Now we can evaluate this expression:
F = 119.0 N + 194.208 N
Adding the two values together gives:
F ≈ 313.208 N
Therefore, the net force on the rocket 3.90s after fuel ignition is approximately 313.208 N.
Now, let's move on to finding the acceleration of the rocket 3.90s after fuel ignition.
The equation that relates force, mass, and acceleration is:
F = ma
To find the acceleration, we rearrange the equation:
a = F/m
Given:
Force, F = 313.208 N (calculated in Part C)
Mass, m = 6.90 kg (given)
Substituting these values into the equation:
a = 313.208 N / 6.90 kg
Calculating this expression gives:
a ≈ 45.33 m/s^2
Therefore, the acceleration of the rocket 3.90s after fuel ignition is approximately 45.33 m/s^2.