Complete Question:
During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N/ Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass, which is 62 kg.
a) What is the minimum stopping time interval in needed to avoid injury to the knee if the person is initially traveling at 15 m/s (34 mi/h)? b) what is the minimum stopping distance?
Answer:
a) t= 46.5 msec. b) 0.35 m
Explanation:
Applying Newton´s 2nd Law to the mass supported by the knee (20% of the total mass), we can get the maximum acceleration allowable in order to avoid an injury, as follows:
a = F/m = 4000 N / 0.2*62 kg = 322.6 m/s²
Applying the definition of acceleration, and taking into account that the knee finally come to an stop, we have:
a = vf - v₀ / Δt = -15 m/s / Δt
Solving for Δt :
Δt = -15 m/s / -322.6 m/s² = 0.0465 sec = 46.5 msec.
b) Assuming the acceleration remains constant during this time interval, we can find the distance needed to come to an ⇒stop, applying any of the kinematic equations, as this one:
vf² - v₀² = 2*a*Δx
⇒Δx = (vf²-v₀²) / 2*a
⇒Δx = -(15 m/s)² / 2*(-322.6 m/s²) = 0.35 m
This question involves the concepts of the equations of motion and Newton's Second Law of Motion.
a) The minimum stopping time interval needed to avoid knee injury is "0.05 s".
b) The minimum stopping distance is "0.37 m".
a)
First, we will use Newton's Second Law of Motion to find out the acceleration:
[tex]F=ma\\\\a=\frac{F}{m}[/tex]
where,
a = acceleration = ?
m = mass supported by knee = 20% of total mass = (0.2)(65 kg) = 13 kg
Assuming the average mass of a person to be 65 kg.
Force = - 4000 N (reaction force)
Therefore,
[tex]a=\frac{-4000\ N}{13\ kg}\\\\a=-307.7\ m/s^2[/tex] (negative sign shows decelration)
Now, for the minimum time interval, we will use the first equation of motion:
[tex]v_f=v_i+at\\\\t=\frac{v_f-v_i}{a}[/tex]
where,
t = time interval = ?
vf = final speed = 0 m/s
vi = initial speed = 15 m/s
Therefore,
[tex]t=\frac{0\ m/s-15\ m/s}{-307.7\ m/s^2}[/tex]
t = 0.05 s = 50 ms
b)
Now, we will use the second equation of motion to find out the stopping distance:
[tex]s=v_it+\frac{1}{2}at^2\\s=(15\ m/s)(0.05\ s)+\frac{1}{2}(-307.7\ m/s^2)(0.05\ s)^2\\[/tex]
s = 0.75 m - 0.38 m
s = 0.37 m
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion.
