Question

a motor car of mass 1200 kg slows down from 72km/h to 36km/h over at a distance of 25 m. If the brakes are applied with the same force. Calculate - (a) total time in which car comes to rest, (b) distance travelled by it, (c) magnitude of force

Answer 1

mass = 1200 kg
u = 72 km/h = 20 m/s
v = 36 km/h = 10 m/s 
s = 25 m

From 3rd equation of motion, we get :
2as = v² - u²
2(a)(25) = (10)² - (20)²
50 a = 100 - 400 

50 a = -300

a = -300 / 50 = -6 m/s²  { - because a is in opposite direction }

Now in our case, (b)
u = 10 m/s
v = 0 
a = -6 m/s²
2as = v² - u²
2(-6)(s) = 0 - 100
-12s = -400
s = 33.33 m 

In case (a)
v = u + at
0 = 10 + (-6)t
0 = 10 - 6t
-6t = -10
t = 1.66 s

F = m * a
= 1200 * (-6) = - 7200N  { - because it is in opposite direction }