A 33300 lb (earth weight) rocket in outer space has two constant forces acting upon it,
F1 = (-166i + 392j) lb and
F2 = (50i + -54j) lb.
Its initial velocity is (39i + 64j) ft/s.
How far is it from where it started after 33 seconds?
To find the distance the rocket travels after 33 seconds, we can use the kinematic equation:
d = v0t + (1/2)at^2
where:
d is the distance traveled,
v0 is the initial velocity,
t is the time, and
a is the acceleration.
First, let's find the acceleration of the rocket. Since the rocket is in outer space, the only forces acting on it are F1 and F2. The total force acting on the rocket can be found by adding the individual forces:
Ftotal = F1 + F2
Ftotal = (-166i + 392j) + (50i - 54j)
= (-166i + 50i) + (392j - 54j)
= (-116i + 338j) lb
Next, we need to find the acceleration of the rocket using Newton's Second Law:
Ftotal = mass * acceleration
Since the mass of the rocket is not given, we will assume it to be m lb:
(-116i + 338j) = m * acceleration
Now, let's use the given information about the rocket's weight to find its mass. The weight of the rocket on Earth can be calculated using the formula:
Weight = mass * gravitational acceleration
where the gravitational acceleration on Earth is approximately 32.2 ft/s^2. So,
33300 lb = m * 32.2 ft/s^2
Dividing both sides by 32.2 ft/s^2, we get:
m = 33300 lb / 32.2 ft/s^2
m ≈ 1033.23 lb·s^2/ft
Substituting the value of m back into the equation for Ftotal, we have:
(-116i + 338j) = (1033.23 lb·s^2/ft) * acceleration
Now we can find the acceleration:
acceleration = (-116i + 338j) / (1033.23 lb·s^2/ft)
acceleration ≈ (-0.1122i + 0.3269j) ft/s^2
Finally, we can substitute the values of the initial velocity (v0) and the time (t) into the kinematic equation to find the distance (d) traveled by the rocket:
d = (39i + 64j) ft/s * 33 s + (1/2) * (-0.1122i + 0.3269j) ft/s^2 * (33 s)^2
Simplifying the equation, we have:
d ≈ (1287i + 2112j) ft + (-0.0561i + 0.5396j) ft
Adding the like terms, we get:
d ≈ (1286.9439i + 2112.5396j) ft
Therefore, the rocket is approximately 1286.94 ft to the right and 2112.54 ft above its starting position after 33 seconds.