Question

A 3000 kg truck moving at a speed of 90m/s stops after covering some distance. The force applied by brakes is 27000N. Compute the distance covered and work done by this force

Answer 1

a = -27000/3000 = -9 m/s^2.
u = 90m/s.
v = 0 m/s
Using, v^2 - u^2 = 2as
s = - 90 × 90 / 2 × -9
s = 450 m.

P.s. The negative sign under force, and acc. insicates that they are in a direction opposite to motion.

Answer 2

deaccleration=27000/3000
v^2 =u^2 +2as
0=u^2-2as
u=90m/s
90×90=2×27000/3000×s
s=8100÷18
s=450m