Question

A train is travelling at a speed of 60 km/h.brakes are applyed as to produce a uniform acceleration of -0.5 m/s² . find how far will the train go before it is brought to rest .​

Answer 1

Given

☞ $\:\begin{cases}\sf\red{Initial\: speed \;of \:train\: =\: 60 km/h}\\\sf\orange{Acceleration\: =\: -0.5 m/s^2}\\\sf\blue{Train \ is \ brought \ to \ rest}\end{cases}$

To find

• $\bf\blue{Distance}$ covered by train

Solution

${\small{\bf{We\:Have}}}\:\begin{cases}\sf\red{Initial\: speed \:(u)\: =\:16.67 m/s\:\bigg\{60 \times\dfrac{ 5}{18}\bigg\}}\\\sf\orange{Acceleration \ (a) \ = \ -0.5 m/s^2}\\\sf\green{Final \ speed \ (v) \ = \ 0 m/s}\end{cases}$

Using 3rd equation of motion

$\to\:\:\sf{ v^2 - u^2 = 2as}$

$\to\:\:\sf{ 0^2 - (16.67)^2 = 2 \times (-0.5)s}$

$\to\:\:\sf{ 0 - 277.89 = -s}$

$\to\:\:\sf{ -277.89 = -s}$

$\to\:\:\sf{ s = {\bf{\pink{277.89 m}}}}$

Hence,

• Distance covered by train before coming to rest = $\bf\red{277.89 m}$

Answer 2

Answer:

Initialspeedoftrain=60km/h

Acceleration=−0.5m/s

2

Train is brought to rest

To find

\bf\blue{Distance}Distance covered by train

Solution

\begin{gathered}{\small{\bf{We\:Have}}}\:\begin{cases}\sf\red{Initial\: speed \:(u)\: =\:16.67 m/s\:\bigg\{60 \times\dfrac{ 5}{18}\bigg\}}\\\sf\orange{Acceleration \ (a) \ = \ -0.5 m/s^2}\\\sf\green{Final \ speed \ (v) \ = \ 0 m/s}\end{cases}\end{gathered}

WeHave

⎩

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎧

Initialspeed(u)=16.67m/s{60×

18

5

}

Acceleration (a) = −0.5m/s

2

Final speed (v) = 0m/s

Using 3rd equation of motion

\to\:\:\sf{ v^2 - u^2 = 2as}→v

2

−u

2

=2as

\to\:\:\sf{ 0^2 - (16.67)^2 = 2 \times (-0.5)s}→0

2

−(16.67)

2

=2×(−0.5)s

\to\:\:\sf{ 0 - 277.89 = -s}→0−277.89=−s

\to\:\:\sf{ -277.89 = -s}→−277.89=−s

\to\:\:\sf{ s = {\bf{\pink{277.89 m}}}}→s=277.89m