A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 7.10 later. What was the rocket's acceleration?
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To find the rocket's acceleration, we can use the kinematic equation that relates displacement, initial velocity, time, and acceleration:
s = ut + (1/2)at^2
Where:
s = displacement (distance)
u = initial velocity
t = time
a = acceleration
In this case, we have the values for time (t) and the total displacement (s) of the bolt. The total displacement of the bolt is the distance it has fallen, which is equivalent to the height it would have reached if it remained on the rocket.
According to the problem, the bolt fell off the rocket 4 seconds after liftoff and hit the ground 7.10 seconds later. Therefore, the time it took for the bolt to hit the ground is given by:
t = 4 + 7.10
t = 11.10 seconds
We also know that the displacement of the bolt is equal to the height it would have reached if it remained on the rocket, which we'll denote as s_bolt. Therefore, the displacement of the bolt is given by:
s = s_bolt
Now, let's calculate the displacement of the bolt using the equation:
s = ut + (1/2)at^2
We can assume that the initial velocity of the bolt when it fell off is 0 since it was initially at rest relative to the rocket:
s = (1/2)at^2
Plugging in the values we know:
s_bolt = (1/2)a(11.10)^2
Now, we can solve for the acceleration (a):
a = (2s_bolt) / (11.1^2)
Substituting the given values, with s_bolt being the displacement of the bolt:
a = (2 * s_bolt) / (11.1^2)
By calculating (2 * s_bolt) / (11.1^2), we can find the rocket's acceleration.