highway saftey engineers build soft barriers so that cars hittingthem will slow down at a safe rate. a person wearing a saftey belt can withstand an acceleration of 300 m/s how thick should barriers be to safely stop a car that hits a barrier at 110 km/h
The deceleration sustained by the safety belt is 300m/s²
110 km/h = 30.55 m/s
The distance travelled (through the soft barrier) while stopping the vehicle at a deceleration of 30.6 g is
(v⊃-u²)/2a
=(0-30.55²)/(2*300)
=1.56 m
To determine the thickness of the soft barriers required to safely stop a car hitting them at a given speed, several factors need to be considered: the initial velocity of the car, the maximum acceleration tolerated by a person wearing a safety belt, and the distance required to bring the vehicle to a stop.
First, we need to convert the given velocity from km/h to m/s.
Given:
Initial velocity (v) = 110 km/h
To convert km/h to m/s, we use the conversion factor: 1 km/h = 1/3.6 m/s.
110 km/h = (110/3.6) m/s = 30.56 m/s (rounded to two decimal places)
Next, we need to determine the stopping distance required to bring the car to a safe stop. This can be calculated using the equation of motion:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity = 0 m/s (since the car needs to come to a stop)
vi = initial velocity = 30.56 m/s
a = acceleration
d = stopping distance (what we're trying to find)
Rearranging the equation, we have:
d = (vf^2 - vi^2) / (2a)
Since we know the final velocity (vf) is 0 m/s, and the maximum acceleration (a) a person can withstand is given as 300 m/s^2, we can plug in these values:
d = (0^2 - 30.56^2) / (2 * 300)
Simplifying further:
d = -(30.56^2) / (2 * 300)
d ≈ -315.65 m^2/s^2 (rounded to two decimal places)
The negative sign arises because the acceleration is opposing the motion of the car.
To eliminate the negative sign, we take the absolute value of the stopping distance:
stopping distance (d) ≈ 315.65 m^2/s^2
Now, in order to determine the thickness of the barriers, we need to know the deceleration the barriers can provide. Assuming they provide a constant deceleration, we can use the following equation:
d = (v^2 - u^2) / (2a)
Where:
d = stopping distance
v = final velocity (0 m/s)
u = initial velocity (30.56 m/s)
a = deceleration provided by the barrier (what we're trying to find)
Plugging in the values:
315.65 = (0^2 - 30.56^2) / (2a)
Simplifying further:
315.65 = -30.56^2 / (2a)
Solving for a, the deceleration:
a = -30.56^2 / (2 * 315.65)
a ≈ -1.488 m/s^2 (rounded to three decimal places)
Again, the negative sign indicates that the deceleration is opposing the motion.
Taking the absolute value of deceleration:
acceleration (a) ≈ 1.488 m/s^2
Now, we can determine the thickness of the barrier required using the equation of motion:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity (0 m/s)
vi = initial velocity (30.56 m/s)
a = acceleration (1.488 m/s^2)
d = thickness of the barrier (what we're trying to find)
Rearranging the equation, we have:
d = (vf^2 - vi^2) / (2a)
Plugging in the values:
d = (0^2 - 30.56^2) / (2 * 1.488)
Calculating further:
d = -30.56^2 / (2 * 1.488)
d ≈ 330.54 m (rounded to two decimal places)
Therefore, the thickness of the barriers should be approximately 330.54 meters. It is worth noting that this calculation assumes a constant deceleration, and in reality, various factors such as friction and other safety mechanisms can affect the stopping distance and barrier thickness.