What Force Is Needed To Bring 1.10\Xd710squared Kg Car Moving At 22 M/S To Stop In 6s

What force is needed to bring 1.10×10squared kg car moving at 22 m/s to stop in 6s

 

Good Day...

Problem:

what force is needed to bring 1.10×10squared kg car moving at 22 m/s to stop in 6s.

Given:

Vi (initial velocity) = 22 m/s

Vf (final velocity ) = 0 m/s (stop)

t (time)        =   6 seconds

a(acceleration) = ? unknown

Solution:

First:

Solve for acceleration with the formula a = (Vf-Vi) / t

a = (Vf-Vi) / t

  = 0 m/s - 22 m/s / 6 s

  = - 22 m/s / 6 s

  = -3.67 m/s^2

Second:

Solve for force with the formula Force = mass × acceleration

Given:

m (mass) = 1.10 × 10^2 kg or equal to 110 kg

a (acceleration) = -3.67 m/s^2

F (force)   = ? unknown

Force = mass × acceleration

           = 110 kg × - 3.67 m/s^2

           = - 403.7 kg.m/s^2

           = - 403.7 N

Answer:

Force is  - 403.7 N

Hope it helps...-)