Vc = 50km/h = 50000m/3600s = 13.89 m/s.
d = (Vf^2-Vo^2)/2a.
d = (0-(13.89)^2) / -7 = 27.56 m.
Yes, it will stop before hitting the ball; but it will be very close!
A child drops a ball onto the road 28 m in front of a car travelling at 50 km/h. If the car can decelerate at a rate of 3.5 m/s2, will it stop before hitting the ball?
1 answer