Paleontologists estimate that if a Tyrannosaurus rex were to trip and fall, it would have experienced a force of approximately 260,000N acting on its torso when it hit the ground.
Assuming the torso has a mass of 3800kg , 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 .)
a=?
Assuming the torso is in free fall for a distance of 1.21m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground?
delta t (change in t)=?
i've tried everything i wanna cry?
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Don't worry, I'm here to help! Let's start by solving the first part of your question - finding the magnitude of the torso's upward acceleration.
To do this, we can use Newton's second law of motion, which states that the force acting on an object is equal to the product of its mass and acceleration: F = m * a.
In this case, the force acting on the torso when it hits the ground is given as 260,000 N, and the mass of the torso is 3800 kg. So we can rearrange the formula to solve for acceleration:
a = F / m
Substituting the given values, we have:
a = 260,000 N / 3800 kg
Calculating this, we find:
a ≈ 68.42 m/s^2
Now let's move on to the second part of your question - finding the time required for the torso to come to rest once it contacts the ground.
Since the torso is in free fall, we can use one of the kinematic equations of motion to solve for time. The equation we'll use is:
s = u * t + (1/2) * a * t^2
Where:
- s is the distance traveled (given as 1.21 m)
- u is the initial velocity (which is 0 m/s since the torso is at rest)
- a is the acceleration (which we just calculated as 68.42 m/s^2)
- t is the time we want to find
Since the torso comes to rest, the final velocity is also 0 m/s. Therefore, we can rewrite the equation as:
s = (1/2) * a * t^2
Rearranging this equation to solve for t, we have:
t = sqrt(2 * s / a)
Substituting the given values, we get:
t = sqrt(2 * 1.21 m / 68.42 m/s^2)
Calculating this, we find:
t ≈ 0.225 s
So the time required for the torso to come to rest once it contacts the ground is approximately 0.225 seconds.
I hope this helps!