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
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?
2 answers
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