Answer :
Answer:
d = 0.38 m
Explanation:
As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:
vf = v₀ -a*t
If vf = 0, we can solve for v₀:
v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s
With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.
Just for simplicity, we can use the following equation:
[tex]vf^{2} -vo^{2} = 2*a*d[/tex]
where vf=0, v₀ =21.2 m/s and a= -588 m/s².
Solving for d:
[tex]d = \frac{-vo^{2}}{2*a} = \frac{(21.2m/s)^{2} }{2*588 m/s2} =0.38 m[/tex]
⇒ d = 0.38 m
Answer:
A person travels 39 cm in coming to a complete stop in 36 ms at a constant acceleration of 60 g.
Explanation:
Hi there!
The equation of position of an object moving in a straight line at constant acceleration is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position at time t.
x0 = initial position.
v0 = initial velocity.
a = acceleration.
t = time.
So, let's see how much distance the person moves inside the car. Let's imagine that the person is initially at rest and suddenly is accelerated at 60 g (60 · 10 m/s² = 600 m/s²). In this case, x0 and v0 = 0 and the traveled distance will be:
x = 1/2 · 600 m/s² · (0.036)²
x = 0.39 m or 39 cm
Here, we have calculated the distance traveled by a person accelerated at 60 g from rest in 36 ms. Notice that the distance is the same if we calculate the traveled distance of a person that is brought to rest in 36 ms with an acceleration of 60 g.
A person travels 39 cm in coming to a complete stop in 36 ms at a constant acceleration of 60 g.