Question

A TRAIN MOVING ALONG A STRAIGHT TRACK, ACCELERATES FROM 36 KM/H TO 72 KM/H IN 5 SECONDS. HOW MUCH DISTANCE DOES IT COVER IN THIS TIME

Answer 1

• 75 metres is the required answer.

Given:-

• Initial velocity ,u = 36km/h
• Final velocity ,v = 72km/h
• Time taken ,t = 5s

To Find:-

• Distance Covered ,S

Solution:-

According to the Question

Firstly we change the unit here into m/s

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Initial velocity ,u = 36

→ u = 36×5/18

→ u = 2×5

→ u = 10m/s

Again,

Final velocity ,v = 72

→ v = 72×5/18

→ v = 4×5

→ v = 20m/s

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Now, calculating the acceleration of the

• a = v-u/t

where,

• v is the final velocity
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

$\longrightarrow$ a = 20-10/5

$\longrightarrow$ a = 10/5

$\longrightarrow$ a = 2m/s²

Using 3rd equation for calculating the distance covered by train.

• v² = u² + 2as

Substitute the value we get

$\longrightarrow$ 20² = 10² + 2×2 × s

$\longrightarrow$ 400 = 100 + 4s

$\longrightarrow$ 400-100 = 4s

$\longrightarrow$ 300 = 4s

$\longrightarrow$ 4s = 300

$\longrightarrow$ s = 300/4

$\longrightarrow$ s = 75 m

• Hence, the distance covered by the train in the given time interval is 75 metres.

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Answer 2

• $\boxed{\sf{\orange{Distance\:covered\:by\:a\:train\:=\:75\:m}}}$

Explanation :

$\underline{\ddag{\underline{\textsf{\textbf{Given}}}}\ddag}\begin{cases} & \sf{Initial\:velocity\;(u) = \bf{36\;km/h}} \\ \\ & \sf{Final\;velocity\;(v) = \bf{72\;km/h}} \\ \\ & \sf{Time\:taken\:(t) = \bf{5\:s}} \end{cases}$

$\underline{\ddag{\underline{\textsf{\textbf{To\:Find}}}}\ddag}$

• Distance covered (s) = ?

$\underline{\ddag{\underline{\textsf{\textbf{Solution}}}}\ddag}$

• Firstly, changing units of initial velocity and final velocity into m/s :

$\leadsto\qquad\sf u = 36\:\times\:\dfrac{5}{18}\:,\:v = 72\:\times\:\dfrac{5}{18}$

$\leadsto\qquad\sf u = \cancel{36}\:\times\:\dfrac{5}{\cancel{18}}\:,\:v = \cancel{72}\:\times\:\dfrac{5}{\cancel{18}}$

$\leadsto\qquad\sf u = 2\:\times\:5\:,\:v = 4\:\times\:5$

$\leadsto\qquad\bf {u = \red{10\:m/s}\:,\:v = \red{20\:m/s}}$

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

• Now, finding it's acceleration :

$\underline{\ddag{\underline{\textsf{\textbf{Using\:Formula}}}}\ddag}$

$\qquad\red\bigstar\:{\underline{\boxed{\bf{\green{a = \dfrac{v - u}{t}}}}}}$

$\qquad \qquad\tiny \bigstar \: {\underline{\frak{\rm{S}\frak{ubstituting\:all\;known\;values\::}}}}$

$:\Rightarrow\qquad\sf a = \dfrac{20 - 10}{5}$

$:\Rightarrow\qquad\sf a = \dfrac{10}{5}$

$:\Rightarrow\qquad\sf a = \dfrac{\cancel{10}}{\cancel{5}}$

$:\Rightarrow\qquad\bf{ a = \red{ 2\:m/s{}^{2}}}$

• Using 3rd eqn of motion for finding distance travelled by a train :

$\qquad\red\bigstar\:{\underline{\boxed{\bf{\green{v^2 = u^2 + 2as}}}}}$

$\qquad \qquad\tiny \bigstar \: {\underline{\frak{\rm{S}\frak{ubstituting\:all\;known\;values\::}}}}$

$\longmapsto\qquad\sf (20)^2 = (10)^2 + 2\:\times\:2\:\times\:s$

$\longmapsto\qquad\sf 400 = 100 + 4s$

$\longmapsto\qquad\sf 400 - 100 = 4s$

$\longmapsto\qquad\sf 300 = 4s$

$\longmapsto\qquad\sf \dfrac{300}{4} = s$

$\longmapsto\qquad\sf \dfrac{\cancel{300}}{\cancel{4}} = s$

$\longmapsto\qquad\bf{s = \red{75\:m}}$

$\small\red{\therefore}\:{\underline{\sf{Hence,\:distance\:covered\:by\:a\:train\:=\:\bf{75\:m}\:\sf{respectively.}}}}$

$\underline{\ddag{\underline{\textsf{\textbf{More\:to\:know}}}}\ddag}$

• Three equations of motion :

1. v = u + at
2. s = ut + (1/2) at²
3. v² = u² + 2as ㅤㅤㅤ[Used above]

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