Question

A train was moving at the rate of 54 km h-1
when brakes were applied. It came to rest
within a distance of 225 m. Calculate the
retardation produced in the train

Answer 1

here u = 54 km/h = 15 m/s

v = 0

s 225m

V square = u square + 2as

0=225+450a

-225 = 450a

-225/450= a

-0.5m/s square is the retradation

Answer 2

Given :

Initial Velocity of the train (u) = 54 Km/hr

Distance travelled (s) = 225

To Find :

Retardation produced in the train.

Solution :

Using Kinematic equation for uniformly accelerated motion, we have

 v²  =  u²  - 2as            

where, v = final velocity; u = initial velocity; a = retardation; s = distance travelled.

As initial velocity is given in Km/hr we need to first covert it into m/sec

Initial velocity (u) = 54 Km/hr = 54 × $\frac{5}{18}$ = 15 m/sec

∴    v²        =  u²  - 2as  

⇒  0²        =  (15)² - 2×a×225

⇒  0         =  225  - 450a

⇒ 450a   =  225

∴        a   = 0.5 m/sec²

∴ The retardation produced in the train is 0.5 m/sec²