What will the stopping distance be for a 3,000-kg car if -3,000 n of force are applied when the car is traveling 10 m/s? a. 30 m b. 150,000 m c. 50 m d. 10 m

F = force applied to stop the car = - 3000 N

m = mass of the car = 3000 kg

a = acceleration of the car = ?

v₀ = initial velocity of the car before the force is applied to stop it = 10 m/s

v = final velocity of the car when it comes to stop = 0 m/s

d = stopping distance of the car

acceleration of the car is given as

a = F/m

inserting the values

a = - 3000/3000

a = - 1 m/s²

using the kinematics equation

v² = v²₀ + 2 a d

inserting the values

0² = 10² + 2 (-1) d

0 = 100 - 2 d

2 d = 100

d = 100/2

d = 50 m

hence the correct choice is

C. 50 m

come closer and become far apart and become steam/gas

ingenuity says that the answer is 2.7 m.