Question

A train moving with a velocity of 48 km h-' is brought to
rest in 11 second to avoid collision. What is the
retardation of the train in ms-2 and km h-2?

Answer 1

Explanation:

Calculating retardation in m/s² :

• Initial velocity (u) = 48 km/h ⇒ 13.33 m/s

$\longmapsto \rm { u = 48 \; kmh^{-1}}$

$\longmapsto \rm { u = \Bigg ( 48 \times \dfrac{5}{18} \Bigg ) \; ms^{-1}}$

$\longmapsto \rm { u = \Bigg ( 8 \times \dfrac{5}{3} \Bigg ) \; ms^{-1}}$

$\longmapsto \rm { u = \Bigg ( \dfrac{40}{3} \Bigg ) \; ms^{-1}}$

$\longmapsto \bf { u = 13.33 \; ms^{-1}}$

• Final velocity (v) = 0 m/s (as it stops)
• Time taken (t) = 11 seconds

In order to calculate retardation, we need to calculate acceleration first.

By using the first equation of motion :

$\longmapsto \bf{v = u + at }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\longmapsto \rm{0 =13.33+ 11a }$

$\longmapsto \rm{0 -13.33 = 11a }$

$\longmapsto \rm{-13.33 = 11a }$

$\longmapsto \rm{\cancel{\dfrac{ -13.33}{11}} = a }$

$\longmapsto \rm{ -1.21 \; ms^{-2} = a }$

Now, as the acceleration is negative, so the retardation will be positive.

$\longmapsto \bf{ 1.21 \; ms^{-2} = Retardation }$

∴ Retardation is 1.21 m/s².

$\rule{200}2$

Calculating retardation in km/h² :

• Initial velocity (u) = 48 km/h
• Final velocity (v) = 0 km/h (as it comes to rest)
• Time taken (t) = 11 seconds

$\longmapsto \rm { t = 11 \;s }$

$\longmapsto \rm { t =\Bigg( 11 \times \dfrac{1}{360} \Bigg ) \;h }$

$\longmapsto \bf { t =\dfrac{11}{360} \;h }$

By using the first equation of motion :

$\longmapsto \rm{0 = 48+ \dfrac{11}{360}a }$

$\longmapsto \rm{0 -48= \dfrac{11}{360}a }$

$\longmapsto \rm{-48(360) = 11a }$

$\longmapsto \rm{-17280 = 11a }$

$\longmapsto \rm{\cancel{\dfrac{ -17280}{11}} = a }$

$\longmapsto \rm{ -1570.909 \; kmh^{-2} = a }$

Now, as the acceleration is negative, so the retardation will be positive.

$\longmapsto \bf{ 1570.909 \; kmh^{-2} = Retardation }$

∴ Retardation is 1570.909 km/h².

Answer 2

Answer:

u = 48 km/hr

t = 11 sec

= 11 / (60×60) hr

v = 0 km/hr

By Newton's first law of motion

v = u + at

For retardation

v = u - at

0 = 48 - a × 11/60 × 60

a = (48 × 60 × 60)/11

a ≈ 15709 km/hr