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 brakes pedal. What must the car’s deceleration if the car is to stop just
before it reaches the child?
at deceleration a, the velocity is
v = 20-at
so t = 20/a
the distance traveled by the car is
s = 20t + 1/2 at^2
= 20(20/a) - 1/2 a (20/a)^2
= 200/a
So, to stop in 50 meters,
200/a = 50
a = 4 m/s^2
check:
v = 20-4*5
s = 20*5 - 2*5^2 = 100-50 = 50
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To find the car's deceleration, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (which is zero because the car is stopping)
u = initial velocity of the car (20 m/s)
s = distance traveled by the car (50 m)
a = acceleration (but since the car is decelerating, it's the negative value of acceleration)
Substituting the values into the equation, we get:
0^2 = 20^2 + 2*(-a)*50
Simplifying the equation:
0 = 400 - 100a
Rearranging the equation to solve for acceleration (-a):
100a = 400
a = 400/100
a = 4 m/s^2
Therefore, the car's deceleration must be 4 m/s^2 for it to stop just before it reaches the child.