Problem:
A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 "g's.
Part A:
Calculate the magnitude of the force on a 71-kg person accelerating at this rate.
Answer=> F= 2.09x10^4 N
Part B:
What distance is traveled if brought to rest at this rate from 95 km/h ?
Answer=> d=______m
Part B:
Oh, I see we're going for a crash course in physics! Buckle up, because we're about to calculate some distances!
First, let's convert the speed from km/h to m/s. We all know how much scientists love their units!
95 km/h = 26.39 m/s (approximately)
Now, we need to use the formula for distance traveled during deceleration:
d = (v^2 - u^2) / (2a)
Where:
- v = final velocity (0 m/s, because we're bringing this person to a stop)
- u = initial velocity (26.39 m/s)
- a = deceleration (-30 g's, which is approximately -294 m/s^2... that's some serious braking power!)
Plugging in the values and crunching through the math:
d = (0^2 - 26.39^2) / (2 * -294)
= (-696.4321) / (-588)
≈ 1.18 meters
So, it looks like the person would travel approximately 1.18 meters before coming to a stop. That's not too far, but still, let's be extra careful out there on the roads! Safety first!
To solve Part B of the problem, we need to calculate the distance traveled when a person brought to rest at a deceleration rate of 30 "g's from a velocity of 95 km/h.
Step 1: Convert the initial velocity to m/s.
1 km/h = 1000 m/3600 s = 0.2778 m/s
The initial velocity is given as 95 km/h:
v = 95 km/h * 0.2778 m/s = 26.39 m/s
Step 2: Determine the final velocity.
In this case, the person is brought to rest, so the final velocity (vf) is 0 m/s.
Step 3: Use the following equation to calculate the distance (d):
vf^2 = vi^2 + 2ad
Since vf = 0 m/s, the equation becomes:
0 = (26.39 m/s)^2 + 2 * a * d
Simplifying the equation:
Distribute the 2 to the terms on the right side of the equation:
0 = 694.6521 m^2/s^2 + 2ad
Rearranging the equation:
-694.6521 m^2/s^2 = 2ad
Step 4: Solve for d.
Divide both sides of the equation by 2a:
d = -694.6521 m^2/s^2 / (2a)
Given that the deceleration rate is 30 "g's, which can be converted to m/s^2 as follows:
1 g = 9.8 m/s^2
30 g's = 30 * 9.8 m/s^2 = 294 m/s^2
Substitute the values into the equation:
d = -694.6521 m^2/s^2 / (2 * 294 m/s^2)
Simplify the equation:
d = -694.6521 m^2/s^2 / 588 m/s^2
Finally, calculate the distance:
d = -1.181 m^2/s^2
However, distance cannot be negative in this context. So, the answer is not physically meaningful and needs to be discarded.
To calculate the magnitude of the force on a person accelerating at a rate of 30 "g's," we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
Given:
Mass (m) = 71 kg
Acceleration (a) = 30 "g's"
First, we need to convert the acceleration from "g's" to meters per second squared (m/s^2), as it is a more standard unit in the metric system.
1 "g" is equal to the acceleration due to gravity on Earth, which is approximately 9.8 m/s^2.
So to convert 30 "g's" to m/s^2, we multiply it by 9.8:
Acceleration (a) = 30 * 9.8 m/s^2 = 294 m/s^2
Now we can use Newton's second law to calculate the force:
Force (F) = Mass (m) * Acceleration (a)
F = 71 kg * 294 m/s^2
F = 2.09 * 10^4 N
Therefore, the magnitude of the force on the 71-kg person accelerating at a rate of 30 "g's" is 2.09 * 10^4 N.
Moving on to Part B, to calculate the distance traveled when brought to a rest at an acceleration of 30 "g's" from a speed of 95 km/h, we need to use the equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity (which is 0 when brought to rest),
- u is the initial velocity (95 km/h),
- a is the acceleration (-294 m/s^2, given that the person is decelerating),
- s is the distance traveled.
First, we need to convert the initial velocity from km/h to m/s:
1 km/h is equal to 0.2778 m/s.
So, the initial velocity (u) in m/s is:
u = 95 km/h * 0.2778 m/s = 26.39 m/s
Now, we can solve for the distance (s) using the equation:
0^2 = (26.39)^2 + 2 * (-294) * s
Simplifying further:
0 = 695.52 - 588s
Rearranging the equation:
-695.52 = -588s
Finally, solving for s:
s = -695.52 / -588
s ≈ 1.183 meters
Therefore, the distance traveled when brought to rest at an acceleration of 30 "g's" from a speed of 95 km/h is approximately 1.183 meters.
A. F = M*g*30 = 71*9.8*30 = 20,874 N. = 2.09*10^4 N.
B. Vo = 95km/h = 95,000m/3600s = 26.4 m/s.
F = M*a = -20,874. a = -20,873/71 = -294 m/s^2.
V^2 = Vo^2 + 2a*d. V = 0, Vo = 26.4 m/s, a = -294 m/s^2, d = ?.