Question

A train is travelling at a speed of 72kmph. The driver applies brakes so that a uniform acceleration of -0.2 ms^-2 is produced. The distance travelled by train before it comes to rest is
a) I km b) 10km c)0.1km d)0.01km

Answer 1

Answer:

Option A

Explanation:

As per the provided information in the given question, we have :.

• Initial velocity (u) = 72 km/h
• Final velocity (v) = 0 m/s [As it stops]
• Acceleration (a) = – 0.2 m/s²

We've been asked to calculate the distance travelled.

In order to calculate the distance covered, firstly we need to convert initial velocity into its standard form.

→ u = 72 km/h

• To convert km/h to m/s, we multiply the value by 5/18.

→ u = (72 × 5/18) m/s

→ u = (4 × 5/1) m/s

→ u = 20 m/s

By using the third equation of motion,

→ v² – u² = 2as

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

→ (0)² – (20)² = 2 × (-0.2) × s

→ 0 – 400 = –0.4 × s

→ – 400 = –0.4s

→ – 400 ÷ (– 0.4) = s

→ 1000 m = s

100 m can also be written as 1 km.

Therefore, the distance travelled by train before it comes to rest is 1 km.

━━━━━━━━━━━━━━━━━━━━━━━━━━

Learn More!

First equation of motion :

• v = u + at

Second equation of motion :

• s = ut + ½at²

Third equation of motion :

• v² = u² + 2as

Where,

• v denotes final velocity

• u denotes initial velocity

• a denotes acceleration

• s denotes distance

• t denotes time

Answer 2

$\huge\purple{ \underline{ \boxed{\mathbb{\red{QUESTION : }}}}}$

A train is travelling at a speed of 72 Kmph . The driver applies brakes so that a uniform acceleration of -0.2 ms^-2 is produced . The distance travelled by train before it comes to rest is :

• a) 1 Km

• b) 10 Km

• c) 0.1 Km

• d) 0.01 Km

$\huge\purple{ \underline{ \boxed{\mathbb{\red{ANSWER : }}}}}$

• a) 1 Km ✔️

$\huge\purple{ \underline{ \boxed{\mathbb{\red{EXPLANATION : }}}}}$

$\green{ \large \underline{ \mathbb{\underline{GIVEN : }}}}$

• Initial Velocity ( u ) = 72 $\sf Km {h}^{ - 1}$

• Acceleration ( a ) = -0.2 $\sf m {s}^{ - 2}$

• Final Velocity ( v ) = 0 $\sf m {s}^{ - 1}$ ( At rest )

$\green{ \large \underline{ \mathbb{\underline{TO \: FIND : }}}}$

• Distance travelled by train ( s ) .

$\green{ \large \underline{ \mathbb{\underline{ EQUATION\: OF \: MOTION \: USED : }}}}$

$\purple{ \boxed{ \sf {v}^{2} - {u}^{2} = 2as }}$

$\green{ \large \underline{ \mathbb{\underline{SOLUTION: }}}}$

$\red{\textsf{ \underline{\underline{Converting initial velocity into SI unit : }}}}$

To convert kilometre per hour into metre per hour , we multiply the value by 5/18 .

$\displaystyle \sf \longrightarrow u = \bigg( 72 \times \frac{5}{18} \bigg)m {s}^{ - 1} \: \\ \\ \displaystyle \sf \longrightarrow u = \bigg( 4 \times 5 \bigg)m {s}^{ - 1} \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow u = 20 \: m {s}^{ - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\red{\textsf{ \underline{\underline{Distance travelled by train ( s ) : }}}}$

$\displaystyle \sf \longrightarrow {v}^{2} - {u}^{2} = 2as \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow {(0)}^{2} - {(20)}^{2} = 2 \times ( - 0.2) \times s \: \\ \\ \displaystyle \sf \longrightarrow - 400 = - 0.4s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow - 0.4s = - 400 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = \frac{ \cancel{-} 400}{ \cancel{-} 0.4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = 1000 \: m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow s = 1 \: Km \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\purple{ \boxed{ \begin{array}{l} \textsf{Distance travelled by train ( s ) = 1 Km} \\ \\ \therefore \textsf{Option ( A ) is correct .}\end{array}}}$

$\green{ \large \underline{ \mathbb{\underline{KNOW\:MORE: }}}}$

• Acceleration

Acceleration is the rate at which velocity changes with time . Negative acceleration is called retardation .

• Initial Velocity

Initial velocity is the velocity of the object before the effect of acceleration .

• Final Velocity

Final velocity is the velocity of the object after the effect of acceleration .

• Distance

Distance is the length of actual path covered by a moving object in a given time interval .

$\Large{\purple{\boxed{\begin{array}{l} \textsf{Equations of motion : } \\ \\ \textsf{v = u + at} \\ \\ \displaystyle\textsf{s = ut + \sf\frac{1}{2}a {t}^{2} } \\ \\ \sf {v}^{2} - {u}^{2} = 2as \end{array}}}}$