Question

A train moving with certain speed acquired a velocity of 600m/s in 5 sec and travels a distance of 1525 m. find its initial velocity .

Answer 1

Given:

Final Velocity,v= 600m/s

Time taken,t= 5sec

Distance travelled,s= 1525m

To Find:

Initial velocity of train

Solution:

Let the initial velocity of train be 'u m/s'

We know that,

➠ According to first equation of motion

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

➠ According to third equation of motion

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

where

• u is initial velocity

• v is final velocity

• a is acceleration

• s is distance covered

• t is time taken

Now, on applying first equation of motion on train, we get

$\longrightarrow\mathrm{600=u+a(5)}$

$\longrightarrow\mathrm{600-u=5a}$

$\longrightarrow\mathrm{5a=600-u}$

$\longrightarrow\mathrm{a=\dfrac{600-u}{5}}$

Now, on applying third equation of motion on train, we get

$\longrightarrow\mathrm{(600)^{2}-u^{2}=2a(1525)}$

On feeding the value of a, we get

$\longrightarrow\mathrm{(600)^{2}-u^{2}=2\bigg(\dfrac{600-u}{\cancel{5}}\bigg)(\cancel{1525})}$

$\longrightarrow\mathrm{360000-u^{2}=2(600-u)(305)}$

$\longrightarrow\mathrm{360000-u^{2}=610(600-u)}$

$\longrightarrow\mathrm{360000-u^{2}=366000-610u}$

$\longrightarrow\mathrm{366000-610u=360000-u^{2}}$

$\longrightarrow\mathrm{u^{2}+366000-610u-360000=0}$

$\longrightarrow\mathrm{u^{2}-610u+6000=0}$

$\longrightarrow\mathrm{u^{2}-600u-10u+6000=0}$

$\longrightarrow\mathrm{u(u-600)-10(u-600)=0}$

$\longrightarrow\mathrm{(u-10)(u-600)=0}$

$\longrightarrow\mathrm{u=10,600}$

Since, 600 m/s is the final velocity.

Therefore, 10 m/s is the initial velocity

Hence, the value of initial velocity of train is 10 m/s.