Question:

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

Solution: Calculating the retardation

Initial velocity(u): (Conversion from Kmh^-1 to ms^-1)

$${54 \ \frac {km}{h}=54 \ (\frac{5}{18}) \ \frac{m}{s}}$$

$${u=15 \ \frac{m}{s}}$$

Final velocity (v):

$${v=0}$$

Brakes were applied to stop the train.

Stopping distance (s):

$${s=225 \ m}$$

Retardation (-a):

Retardation is always against the velocity of the body.

$${v^2=u^2+2as}$$

$${a=\frac{v^2-u^2}{2s}}$$

$${acceleration=\frac{0-15^2}{2(225)}}$$

$${a=\frac{-225}{450}}$$

$${a=\frac{-1}{2}=-0.5 \ \frac{m}{s^2}}$$

Negative sign indicates the acceleration is in the direction opposite to that of motion called the retardation.

Understanding retardation:

Understanding retardation and deceleration is important in understanding various fields such as engineering, transportation and physics where controlling the speed and ensuring the safety of moving objects is important.

Concept of retarding force (F):

Retarding force is a force that acts in the direction opposite to the motion of an object, causing it to slow down.Retarding force reduces the kinetic energy of an object.Retarding force is helpful in understanding and analysing the motion of objects.Common examples are friction,air resistance and viscous drag.

Retarding forces are critical in designing the braking system for vehicles,slowing down skydivers for safe landing and in sports itis helpful in designing sports equipments.

Conclusion:

Initial velocity of the train u=15 ms^-1

Final velocity of the train v= 0 ms^-1

Stopping distance s= 225 m

Retardation produced in the train= – 0.5 ms^-2