Question

26. A train started from the rest from
a station and accelerated at 2
m/second square for 10 s. Then, it ran
at constant speed for 30 s and
thereafter it decelerated at 4
m/second square until it stopped at
the next station. The distance
between two stations is
650 m
O
700 m
750 m
800 m​

Answer 1

Answer:

750 metres

Explanation:

Given:

• Initial velocity of the train = u = 0 m/s
• Acceleration of the train = 2m/s²
• Time for which the train accelerated = 10 seconds
• Time for which it ran at a constant speed = 4 seconds
• Decceleration = 4 m/s²

Using first equation of motion:

V=u+at

V=0+2×10

v=20 m/s

Distance covered in this period using third equation of motion:

V²-u²=2as

20²-0²=2×2×s

400-0=4s

400=4s

s=100 metres

Now it runs at a constant speed of 20 m/s for 30 seconds:

Distance = speed×time

Distance = 20×30

Distance = 600 metres

Now:

Initial velocity = 20 m/s

Acceleration = -4 m/s²

Final velocity = v = 0 m/s

Using third equation of motion:

V²-u²=2as

0²-20²=2×-4×s

-400=-8s

s = 50 metres

Total distance = 50+600+100

Total distance = 750 metres

The total distance covered is equal to 750 metres

Answer 2

Answèr :

The distance between two stations is 750m.

Explanatìon

Given :

1. Initial velocity of the train = u = 0 m/s
2. Acceleration of the train = 2m/s²
3. Time for which the train accelerated = 10 seconds
4. Time for which it ran at a constant speed = 4 seconds
5. Decceleration = 4 m/s²

Solútion :

Initially the Train start from rest and travel a distance $\sf\:S_{1}$ in 10 seconds with acceleration 2m/s²

⇒u = 0

t = 10 sec and a =2m/s²

By using equation of motion :

$\sf\:S_{1}=0+\frac{1}{2}\times2\times10^2$

$\implies\sf\:S_{1}=100m$

and , for first 10 sec

v= u+at =0+ 2×10 = 20m/s

Then, train travelled a distance $\sf\:S_{2}$ with constant speed for 30 seconds.

⇒a= 0 ( since speed is constant ) , t = 30 sec

and v = 20 m/s

By using equation of motion:

$\sf\:S_{2}=20\times30+\frac{1}{2}\times0\times3^2$

$\implies\sf\:S_{2}=600m$

Then, train travelled a distance $\sf\:S_{3}$ and stopped  due to retardation 4 m/s.

⇒v= 0 ,u = 20m/s

a = -4m/s²

By using equation of motion :

$\sf\:v^2=u^2+2aS$

$\implies\sf\:0=20^2+2\times-4\times\:S$

$\implies\sf\:S=50m$

Therefore ,

The total distance between two stations

= $\sf\:S_{1}+S_{2}+S_{3}$

= 100+600+50

= 750 m

$\rule{200}2$

Formula used :

⇒Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

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

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

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$