Respuesta :
a = - 8.34 m/sec² ( deceleration or negative)
Equations for UAM ( uniformly accelerated motion) are:
vf = v₀ ± a*t and s = s₀ + v₀*t + (1/2)*a*t²
In our case, the motion is with deceleration, then
vf = v₀ - a*t and s = s₀ + v₀*t - (1/2)*a*t²
working on these equatios we get:
vf = v₀ - a*t (1) s - s₀ = v₀*t - (1/2)*a*t² (2)
v₀ - vf = a*t
t = (v₀ - vf)/a
By substitution of (1) in equation (2)
s - s₀ = v₀ * (v₀ - vf)/a - (1/2) * a* [(v₀ - vf)/a]²
s - s₀ = (v₀² - v₀*vf)/a - (1/2) * a* (1/a²)* (v₀ - vf)²
s - s₀ = 1/a * ( v₀² - v₀*vf ) - 1/a* (1/2) * (v₀ - vf)²
s - s₀ = 1/a* [ ( v₀² - v₀*vf ) - (1/2) * (v₀ - vf)²]
a * (s - s₀ ) = v₀² - v₀*vf - v₀²/2 - vf²/2 + v₀*vf
a * (s - s₀ ) = (1/2) * v₀² - (1/2)*vf²
a * (s - s₀ ) = (1/2) * ( v₀² - vf²)
We find an expression to calculate the minimum deceleration to stop the car in time to avoid crashing
s₀ = 50 meters s = 0 v₀ = 104 Km/h vf = 0
1 Km = 1000 m and 1 h = 3600 sec
v₀ = 104 Km/h = 28.88 m/sec
a = (1/2) [ (28.88)² - 0 ] / 0 - 50
a = - 8.34 m/sec² ( deceleration or negative)
