Question

A train is travelling with velocity of 40 km h-1 i)what should be acceleration on it so that i may reach a point 10 km ahead in 8 minutes??
ii) what will be its velocity on reaching that point

Answer 1

Given:

Initial velocity of train, u= 40 km/h

Distance to be covered by train, s= 10km

Time given, t= 8 min= 2/15 h

To Find:

i) Acceleration required to cover 10 km in 2/15 h

ii) Final velocity at destination point

Solution:

Let the required acceleration of train be 'a' m/s² and final velocity be 'v' m/s

We know that,

• According to first equation of motion for constant acceleration

$\purple{\boxed{\bf{v=u+at}}}$

• According to second equation of motion for constant acceleration

$\pink{\boxed{\bf{s=ut+\frac{1}{2}at^{2}}}}$

i)

On applying second equation of motion for the given train, we get

$\longrightarrow\mathrm{s=ut+\dfrac{1}{2}at^{2}}$

$\longrightarrow\mathrm{10=40\bigg(\dfrac{2}{15}\bigg)+\dfrac{1}{2}a\bigg(\dfrac{2}{15}\bigg)^{2}}$

$\longrightarrow\mathrm{10=\dfrac{16}{3}+\dfrac{1}{2}a\bigg(\dfrac{4}{225}\bigg)}$

$\longrightarrow\mathrm{10-\dfrac{16}{3}=\dfrac{2a}{225}}$

$\longrightarrow\mathrm{\dfrac{30-16}{\cancel{3}}=\dfrac{2a}{\cancel{225}}}$

$\longrightarrow\mathrm{\cancel{14}=\dfrac{\cancel{2}a}{75}}$

$\longrightarrow\mathrm{7=\dfrac{a}{75}}$

$\longrightarrow\mathrm{\dfrac{a}{75}=7}$

$\longrightarrow\mathrm{a=7\times75}$

$\longrightarrow\mathrm{\green{a=525\:km/h^{2}}}$

ii)

On applying first equation of motion for the given train, we get

$\longrightarrow\mathrm{v=u+at}$

$\longrightarrow\mathrm{v=40+\cancel{525}\bigg(\dfrac{2}{\cancel{15}}\bigg)}$

$\longrightarrow\mathrm{v=40+35\times2}$

$\longrightarrow\mathrm{v=40+70}$

$\longrightarrow\mathrm{\blue{v=110\:km/h}}$

Hence, required acceleration is 525 km/h² and final velocity is 110 km/h.

Answer 2

Answer:

a = 525 km/hr² (≈0.04 m/s²)  and  v = 110km/hr (≈30.5 m/s)

Explanation:      Let the required acceleration be 'a'.

In the question,    (unit chosen = km/hr)

   Initial velocity(u) = 40 km/hr

  Time taken(t) = 8min = (2/15) hr

   Displacement(S) = 10 km

                Using equations of motion:

⇒ S = ut + 1/2 at²

⇒ 10 = (40)(2/15) + 1/2 a (2/15)²

⇒ 2250 = 1200 + 2a

⇒ 525 km/hr² = a

     Converting this in m/s², we get

        a = 525 (1000)/(3600)² m/s²

        a ≈ 0.04 m/s²

(ii):  Let the final velocity be 'v'.

⇒ v = u + at

⇒ v = 40 + (525)(2/15)

⇒ v = 110 km/hr

      Converting this in m/s:

        v = 110 (1000)/(3600) m/s

       v ≈ 30.5 m/s

*question is solved with units km and hr  and then converted to m and s.