Respuesta :
A) Final speed: 5.2 m/s
B) TIme to stop: 0.046 s
C) Acceleration: [tex]-113.0 m/s^2[/tex]
D) Acceleration: [tex]-11.5g[/tex]
Explanation:
The complete text of the question is the following:
"A cat drops from a shelf 4.5 ft above the floor and lands on all four feet. His legs bring him to a stop in a distance of 12 cm.
A) Calculate his speed when he first touches the floor (ignore air resistance). Express your answer in SI units.
B) Calculate how long it takes him to stop.
C) Calculate his acceleration (assumed constant) while he is stopping. Express your answer in SI units.
D) Calculate his acceleration (assumed constant) while he is stopping, in g’s."
A)
First of all, we have to convert the initial height of the cat from feet to meters. Since
1 feet = 0.3048 meters
We have
[tex]h=4.5 ft \cdot 0.3048 m/ft=1.37 m[/tex]
The motion of the cat is a free fall, which is a uniformly accelerated motion with constant acceleration [tex]g=9.8 m/s^2[/tex] towards the ground. So we can find the final speed of the cat as it reaches the ground by using the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where, taking downward as positive direction:
v is the final speed
u = 0 is the initial speed
[tex]a=g=9.8 m/s^2[/tex] is the acceleration
s = 1.37 m is the displacement
Solving for v,
[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(1.37)}=5.2 m/s[/tex]
B)
In order to find the time the cat needs to stop, we have to consider the 2nd part of the motion, the one in which it stops in a distance of
s = 12 cm = 0.12 m
This part of the motion is also a uniformly accelerated motion, therefore we can use the following suvat equation:
[tex]s=(\frac{u+v}{2})t[/tex]
where this time we have:
v = 0 is the final speed
u = 5.2 m/s is the initial speed of the cat just before landing
t is the time to stop
s = 0.12 m
And solving for t,
[tex]t=\frac{2s}{u+v}=\frac{2(0.12)}{5.2+0}=0.046 s[/tex]
C)
Here we want to find the acceleration of the cat during the stopping phase.
We can do it by using the following suvat equation:
[tex]v=u+at[/tex]
where:
v = 0 is the final speed
u = 5.2 m/s is the initial speed before the stopping phase
a is the acceleration
t = 0.046 s is the stopping time, found in part B
Solving the equation for a, we find:
[tex]a=\frac{v-u}{t}=\frac{0-5.2}{0.046}=-113.0 m/s^2[/tex]
And this acceleration is negative because the cat is slowing down.
D)
Here we want to find the acceleration of the car expressed in units of g, the gravitational acceleration.
The acceleration of gravity is the value of the acceleration of an object in free fall (=acted upon the force of gravity only). Its value is
[tex]g=9.8 m/s^2[/tex]
Here, the acceleration of the cat during the landing phase is
[tex]a=-113.0 m/s^2[/tex]
Therefore, in order to express this value in units of g, we have to divide the acceleration by the value of g, and we get:
[tex]a=\frac{-113.0}{9.8}=-11.5 g[/tex]
Learn more about accelerated motion:
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