A car is travelling at 20m/s along a road. a child runs out into road 50m ahead and car driver steps on break paddle. what must car retardation rate be if the car is to stop just before it reaches the child?????
V^2 = Vo^2 + 2a*d
a = (V^2-Vo^2)/2d
a = (0-20^2)/100 = -4 m/s^2.
To find the car's required retardation rate, we need to determine the time it takes for the car to stop before reaching the child.
First, let's consider the initial velocity (vᵢ) of the car, which is 20 m/s.
The final velocity (v_f) of the car should be 0 m/s when it stops just before reaching the child.
The car's acceleration (a) can be calculated using the equation:
v_f^2 = vᵢ^2 + 2ad,
where
v_f = final velocity (0 m/s),
vᵢ = initial velocity (20 m/s),
a = acceleration (unknown),
d = distance (50 m).
Substituting the given values into the equation, we get:
0^2 = 20^2 + 2a(50).
Simplifying further:
0 = 400 + 100a.
Rearranging the equation:
100a = -400.
Finally, solving for the acceleration:
a = -400 / 100 = -4 m/s².
Therefore, the car's required retardation rate (acceleration) should be -4 m/s² to stop just before reaching the child.