Paleontologists estimate that if a Tyrannosaurus rex were to trip and fall, it would have experienced a force of approximately 260,000 N acting on its torso when it hit the ground.
Question 1
Assuming the torso has a mass of 3800 kg , find the magnitude of the torso's upward acceleration as it comes to rest. (For comparison, humans lose consciousness with an acceleration of about 7g .)
Question 2
Assuming the torso is in free fall for a distance of 1.21 m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground?
a.
Use F=mA
F=260,000 N
m=3800kg
acceleration A=?
b. v²-v0²=2aS
Initial velocity, v0=0 m/s
acceleration due to gravity, g = -9.8 m/s/s
final velocity (end of free fall), v
distance fallen, S = 1.21 m
what is v?
The moment rex is in contact with the ground, decelleration starts.
Using the upward acceleration from part a., we can apply the formula
v2=v+At
v2=final velocity=0
v=velocity at impact, calculated above.
A=upward acceleration calculated from part a.
t can be isolated and calculated.
To answer both questions, we can use Newton's second law of motion, which states:
F = m * a
where F is the net force acting on an object, m is the mass of the object, and a is the acceleration of the object.
Question 1:
We are given the force acting on the torso, F = 260,000 N, and the mass of the torso, m = 3800 kg. We need to find the upward acceleration (a) of the torso.
Using Newton's second law, we have:
260,000 N = 3800 kg * a
Dividing both sides by 3800 kg:
a = 260000 N / 3800 kg
a = 68.42 m/s^2
Therefore, the magnitude of the torso's upward acceleration as it comes to rest is 68.42 m/s^2.
Question 2:
We are given the distance the torso falls, d = 1.21 m. We need to find the time required for the torso to come to rest once it contacts the ground.
To find the time (t), we can use the kinematic equation:
d = (1/2) * a * t^2
Rearranging the equation:
t^2 = (2 * d) / a
Substituting the values, we can solve for t:
t^2 = (2 * 1.21 m) / 68.42 m/s^2
t^2 = 0.035298 s^2
Taking the square root of both sides:
t = sqrt(0.035298 s^2)
t ≈ 0.188 s
Therefore, it takes approximately 0.188 seconds for the torso to come to rest once it contacts the ground.
To answer both questions, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object times its acceleration.
Question 1:
We are given the mass of the torso (m = 3800 kg) and the force acting on it (F = 260,000 N). We need to find the magnitude of the upward acceleration (a) as it comes to rest. Rearranging the formula, we have:
F = m * a
Solving for a, we get:
a = F / m
Substituting the given values into the equation, we have:
a = 260,000 N / 3800 kg
Calculating this, we find:
a ≈ 68.42 m/s² (rounded to two decimal places)
Therefore, the magnitude of the torso's upward acceleration as it comes to rest is approximately 68.42 m/s².
Question 2:
We are told that the torso falls freely for a distance of 1.21 m. We need to find the time required for the torso to come to rest once it contacts the ground.
We can use the kinematic equation:
s = ut + (1/2)at²
Where:
s = displacement (1.21 m)
u = initial velocity (0 m/s, as it starts from rest)
a = acceleration (-68.42 m/s², negative because it acts in the opposite direction to motion)
t = time (unknown)
Rearranging the equation, we have:
t = sqrt((2s / a))
Substituting the given values into the equation, we have:
t = sqrt((2 * 1.21 m) / (-68.42 m/s²))
Calculating this, we find:
t ≈ sqrt(0.03531) s (rounded to five decimal places)
t ≈ 0.18798 s (rounded to five decimal places)
Therefore, it takes approximately 0.188 seconds for the torso to come to rest once it contacts the ground.