Question

A train is moving with 90km/hrs and break are applied after 50seconds train stop. Calculate acceleration produced and distance travelled by the train before stopping.​

Answer 1

Given :-

Initial velocity (u) = 90km/h

Convert in m/s

$\sf \cancel{90}\times \dfrac{5}{\cancel{18}}= 5\times 5= 25m/s\\ \\ \sf u= 25m/s$

Final velocity (v) =0. [ because break applied ]

Time (t)=50 seconds

To find :-

Acceleration and distance travelled before stopping

• Formula used!!

$\boxed{\textsf{\textbf{\dag \ \ v=u+at}}}$

• Find acceleration (retardation )

$\longmapsto\sf 0= 25+a\times 50\\ \\ \longmapsto\sf -25=50a\\ \\ \longmapsto\sf \cancel{\dfrac{-25}{50}}=a\\ \\ \longmapsto\sf a= \dfrac{-1}{2}= (-0.5m/s^2)$

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

• where

v= final velocity

u=Initial velocity

a= acceleration

s= Distance

• Now the distance

$\longmapsto\sf (0)^2= (25)^2+2\times (-0.5)\times s\\ \\ \longmapsto\sf 0= 625 -s\\ \\ \longmapsto\sf s=625m$

$\bigstar{\boxed{\sf{ Acceleration (retardation)= (-0.5m/s^2)}}}$

$\bigstar{\boxed{\sf{ Distance = 625m}}}$

Answer 2

Answer:

Acceleration = $\frac{-1}{2} m/s^{2}$

Distance = $625 \ metres$

Explanation:

Given:

Initial velocity (u) = 90 km/hr = $90 \times \frac{5}{18}$ = 25 m/s

Time (t) = 50 seconds

Final velocity (v) = 0 m/s (As the train finally comes to rest)

To find:

• Acceleration
• Distance

We can find the Acceleration using first equation of motion, which says:

$V=u+at$

Where:

V= final velocity

u = initial velocity

a = acceleration

t = time

Substituting the above values, we get:

0=25+a×50

-25=50a

a = $\frac{-25}{50}$

a = $\frac{-1}{2}$ m/s²

The acceleration produced by the train is $\frac{-1}{2} m/s^{2}$

We can use the third equation of motion to find the distance which says:

$v^{2} -u^{2} =2as$

Where:

V= final velocity

u = initial velocity

a = acceleration

s = distance

Substituting the above values, we get:

$0^{2}-25^{2} =2 \times \frac{-1}{2} \times s$

$-625 = -1s$

s = $\frac{-625}{-1}$

s = 625 metres

The acceleration produced is $\frac{-1}{2}$ m/s² and the distance is 625 metres