Question

8. A train starting from rest attains a velocity of 90 km/h in 2 min, then the distance travelled by the train for attaining this velocity is (a) 1.5 km (b)2 km (c) 2.5 km (d) 1.2 km​

Answer 1

Answer:

Given :

• Final velocity = 90 km/h
• Initial velocity = 0 km/h
• Time taken = 2 min

To Find :

• Distance travelled by the train

Using Formulas :

First equation of motion :-

$\longrightarrow\underline{\boxed{\sf{v = u + at}}}$

Second equation of motion :-

$\longrightarrow\underline{\boxed{\sf{s = ut + \dfrac{1}{2} \times a{t}^{2} }}}$

Solution :

☼ Coverting the final velocity into m/s :-

$\longrightarrow{\small{\sf{Final \: velocity = 90 \times \dfrac{5}{18}}}}$

$\longrightarrow{\small{\sf{Final \: velocity = \dfrac{90 \times 5}{18}}}}$

$\longrightarrow{\small{\sf{Final \: velocity = \dfrac{450}{18}}}}$

$\longrightarrow{\small{\sf{Final \: velocity = \cancel\dfrac{450}{18}}}}$

$\longrightarrow{\small{\bf{\purple{Final \: velocity = 25 \: m/s}}}}$

∴ The final velocity is 25 m/s.

━┅━┅━┅━┅━┅━┅━┅━┅━┅━

☼ Coverting the Initial velocity into m/s :-

$\longrightarrow{\small{\sf{Initial \: velocity = 0\times \dfrac{5}{18}}}}$

$\longrightarrow{\small{\sf{Initial \: velocity = \dfrac{0 \times 5}{18}}}}$

$\longrightarrow{\small{\sf{Initial \: velocity = \dfrac{0}{18}}}}$

$\longrightarrow{\small{\bf{\purple{Initial \: velocity = 0 \: m/s}}}}$

∴ The initial velocity is 0 m/s.

━┅━┅━┅━┅━┅━┅━┅━┅━┅━

☼ Coverting time taken into sec :-

$\longrightarrow\small{\sf{Time \: taken = 2 \times 60}}$

$\longrightarrow\small{\bf{\purple{Time \: taken = 120\: sec}}}$

∴ The time taken is 120 sec.

━┅━┅━┅━┅━┅━┅━┅━┅━┅━

☼ Now, finding acceleration by using first equation of motion :-

$\longrightarrow\small{\sf{v = u + at}}$

$\longrightarrow\small{\sf{25 = 0+ a \times 120}}$

$\longrightarrow\small{\sf{25 = 0+ 120a}}$

$\longrightarrow\small{\sf{25 = 120a}}$

$\longrightarrow\small{\sf{a = \dfrac{25}{120}}}$

$\longrightarrow\small{\sf{a = \cancel{\dfrac{25}{120}}}}$

$\longrightarrow\small{\bf{\purple{a = {\dfrac{5}{24} \: m/s}}}}$

∴ The acceleration is 5/24 m/s.

━┅━┅━┅━┅━┅━┅━┅━┅━┅━

☼ Now, finding the distance traveled by using third equation of motion :-

$\longrightarrow\small{\sf{s = ut + \dfrac{1}{2} \times a{t}^{2} }}$

${\longrightarrow{\small{\sf{s = \bigg(0 \times 120 \bigg) + \dfrac{1}{2} \times \bigg( \dfrac{5}{24} \times {120}^{2} \bigg) }}}}$

${\longrightarrow{\small{\sf{s = 0 + \dfrac{1}{2} \times \bigg( \dfrac{5}{24} \times 120 \times 120\bigg) }}}}$

${\longrightarrow{\small{\sf{s = 0 + \dfrac{1}{2} \times \bigg( \dfrac{5}{\cancel{24}} \times \cancel{120} \times 120\bigg) }}}}$

${\longrightarrow{\small{\sf{s = 0 + \dfrac{1}{2} \times \big({5 \times 5\times 120\big) }}}}}$

${\longrightarrow{\small{\sf{s = 0 + \dfrac{1}{2} \times 3000}}}}$

${\longrightarrow{\small{\sf{s = 0 + \dfrac{3000}{2}}}}}$

${\longrightarrow{\small{\sf{s = 0 + \cancel{\dfrac{3000}{2}}}}}}$

${\longrightarrow{\small{\sf{s = 0 + 1500 }}}}$

${\longrightarrow{\small{\sf{s =1500 \: m }}}}$

${\longrightarrow{\small{\sf{s = \dfrac{1500}{1000} \: m }}}}$

${\longrightarrow{\small{\sf{s = \cancel{\dfrac{1500}{1000}} \: m }}}}$

${\longrightarrow{\small{\bf{ \purple{s =1.5 \: km }}}}}$

∴ The distance traveled by train is 1.5 km.

━┅━┅━┅━┅━┅━┅━┅━┅━┅━

☼ Note :-

• s = displacement,
• v = Final velocity
• u = Initial velocity
• t = Time

━┅━┅━┅━┅━┅━┅━┅━┅━┅━

Learn More :

$\star$ Velocity

${\longrightarrow{\small{\underline{\boxed{\bf{v = u + at}}}}}}$

$\star$ Displacement with positive accelerations

${\longrightarrow{\small{\underline{\boxed{\bf{s= ut + \dfrac{1}{2}{at}^{2}}}}}}}$

$\star$ Displacement with negative acceleration

${\longrightarrow{\small{\underline{\boxed{\bf{s= ut - \dfrac{1}{2}{at}^{2}}}}}}}$

$\star$ Displacement knowing initial and final speeds

${\longrightarrow{\small{\underline{\boxed{\bf{s= \dfrac{1}{2}\big(u + v \big)t}}}}}}$

$\star$ Velocity squared

${\longrightarrow{\small{\underline{\boxed{\bf{{v}^{2} = {u}^{2} + 2as}}}}}}$

$\underline{\color{gray}{\rule{220pt}{2.5pt}}}$