a typical person can tolerate an acceleration of about -49m/s2 (forward). If you are travelling at 110km/h and have a collison with a solid immovable object, over what minium distance must you stop so as to not exceed this acceleration
To determine the minimum distance required to stop without exceeding the maximum acceleration, we need to convert the speed from km/h to m/s.
Given:
Maximum acceleration (a): -49 m/s^2
Initial speed (u): 110 km/h
Final speed (v): 0 m/s (since the object needs to stop)
Distance (s): ?
First, we convert the initial speed from km/h to m/s:
110 km/h = (110 * 1000) m / (3600 s) ≈ 30.6 m/s
Now, we can use the following equation of motion to find the minimum distance:
v^2 = u^2 + 2as
Plugging in the values:
0 = (30.6 m/s)^2 + 2 * (-49 m/s^2) * s
Rearranging the equation to solve for distance (s):
2 * (-49 m/s^2) * s = - (30.6 m/s)^2
2 * (-49 m/s^2) * s = -938.16 m^2/s^2
-98 m/s^2 * s = -938.16 m^2/s^2
s = (-938.16 m^2/s^2) / (-98 m/s^2)
s ≈ 9.6 m
Therefore, the minimum distance required to stop without exceeding the given acceleration is approximately 9.6 meters.
To determine the minimum distance required to stop without exceeding the tolerable acceleration, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s since you want to come to a stop)
u = initial velocity (110 km/h converted to m/s)
a = acceleration (-49 m/s^2)
s = distance
First, let's convert the initial velocity from km/h to m/s:
110 km/h = (110 * 1000) / (60 * 60) m/s ≈ 30.5556 m/s
Now, let's rearrange the equation to solve for distance (s):
s = (v^2 - u^2) / (2a)
Substituting the values into the equation:
s = (0^2 - (30.5556)^2) / (2 * -49)
Calculating the result:
s = (-931.2304) / (-98)
s ≈ 9.507 m
Therefore, the minimum distance required to stop without exceeding the given acceleration is approximately 9.507 meters.
Vi = 110,000 meters / 3600 seconds
v = Vi + a t
at end, v = 0
so
0 = Vi -49 t
t = Vi/49
then
d = Vi t + (1/2) a t^2
d = Vi (Vi/49) - (1/2)(49)(VI^2/49^2)
d = (1/2)Vi^2/49