A car is travelling at 20m/s along a road. A child runs out into the road 50m ahead and the car driver steps on the brake pedal. What must the car's declerations be if the car is to slop just before it reaches the child .
50 = Vi t + .5 a t^2
v = Vi + a t = 0 at end
so
a = -Vi/t = -20/t
50 = 20 t - .5 (20/t)t^2
50 = 20 t - 10 t = 10 t
t = 5 seconds
a =-20/5 = -4 meters/second^2
-4m/s^2
To determine the car's required deceleration, we need to consider the following:
1. Initial velocity of the car (u): 20 m/s
2. Final velocity of the car (v): 0 m/s (since the car must stop just before it reaches the child)
3. Distance traveled by the car (s): 50 m (distance between the car and the child)
4. Acceleration of the car (a): Unknown, but we can solve for it.
We can use the kinematic equation to find the required deceleration:
v^2 = u^2 + 2as
Substituting the known values into the equation:
0^2 = (20^2) + 2a(50)
Simplifying the equation:
0 = 400 + 100a
Rearranging the equation to solve for a:
100a = -400
a = -400/100
a = -4 m/s^2
Therefore, the car's required deceleration must be -4 m/s^2 (negative sign indicates deceleration) in order to stop just before it reaches the child.