What is the newtons second law of acceleration in the braking distance of vechicles?

Answer 1

install some (arbitrary) numbers:

mass (m) = 1 000 kg

( braking ) force (f) = -10 000 newtons (negative because the force is opposing direction of motion)

braking from 70 metres / sec to rest, so:

u = 70 metres / sec

v = 0 metres / sec

.

first find acceleration (a):

newtons second law: force (f) = mass (m) * acceleration (a)

shuffle to isolate acceleration (a)

acceleration (a) = force / mass = -10 000 / 1 000 = - 10 (m / s) / s

.

now find (braking) distance (s):

use v^2 = (u^2 ) + (2 * a * s)

shuffle to isolate s

s = (v^2 - u^2) / (2 * a)

s = - 4 900 / -20

s = 245 metres

.

time (t) to stop:

use v = u + a * t

shuffle to isolate t

t = (v - u ) / a

t = - 70 / - 10

t = 7 seconds

.

note: these calculations ignore rolling and air resistance, which are additional to the braking force.