The minimum distance required to stop a car moving at 35.6 mi/h is 40.8 ft. what is the minimum stopping distance, in ft, for the same car moving at 71.5 mi/h, assuming the same rate of acceleration? (60.0 mi/h = 88.0 ft/s).

Minimum stopping distance = 164.69 ft

Explanation:

Speed of car = 35.6 mi/hr = 15.82 m/s

Stopping distance = 40.8 ft = 12.44 m

We have equation of motion

             v² = u² + 2as

             0²=15.82²+ 2 x a x 12.44

              a = -10.06 m/s²

Now wee need to find minimum stopping distance, in ft, for the same car moving at 71.5 mi/h.

          Speed of car = 71.5 mi/h = 31.78 m/s

We have

           v² = u² + 2as

          0² = 31.78² -  2 x 10.06 x s

           s = 50.2 m = 164.69 ft

Minimum stopping distance = 164.69 ft

i think it's 35 meters

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j