Question

A train starting from rest attains a velocity of 54 km/h in 5 minutes. Assuming that the acceleration is uniform find the (1) acceleration and the distance travelled by the train for attaining this velocity

Answer 1

GiveN :

• Initial velocity (u) = 0 m/s
• Final velocity (v) = 54 km/h
• Time interval (t) = 5 mins

To FinD :

• Acceleration and distance travelled by train for attaining this velocity.

Formulae UseD :

• v = u + at
• v² - u² = 2as

Where,

• v is final velocity
• u is initial velocity
• a is acceleration
• t is time interval
• s is distance travelled

SolutioN :

First convert velocity and time in SI unit which is m/s and s respectively.

• Final velocity (v) = 54 * 5/18 = 15 m/s
• Time interval (t) = 5 * 60 = 300 s

Acceleration of train :

Use 1st equation of motion :

⇒v = u + at

⇒15 = 0 + a * 300

⇒15 = 300a

⇒a = 15/300

⇒a = 0.05

∴ Acceleration of train is 0.05 m/s²

____________________________

Distance travelled by train

Use 3rd equation of motion :

⇒v² - u² = 2as

⇒15² - 0² = 2 * 0.05 * s

⇒225 - 0 = 0.1*s

⇒225 = 0.1s

⇒s = 225/0.1

⇒s = 225/0.1

⇒s = 2250

∴ Distance travelled by train is 2250 m.

Answer 2

Given :-

A train starts from rest and attains a velocity of 54 km/hr in 5 minutes . The acceleration is constant .

Required to find :-

• Acceleration and distance travelled by the train to attain this velocity

Equations used :-

➜ 1. v = u + at ----|★

➜ 2. v² - u² = 2as -----|★

Solution :-

It is given that ;

A train starts from rest and attains a velocity of 54 km/hr in 5 minutes . The acceleration is constant .

we need to find the acceleration and distance travelled by the train to attain this acceleration .

Since, it is clearly mentioned that the acceleration is constant . So the equations of motion are applicable here .

From the given information we get ;

• Initial velocity = 0 km/hr

• Final velocity = 54 km/hr

• Time = 5 minutes

Before, using the equation we need to convert the speed from km/hrinto m/s and time from minutes to seconds

$\tt \bf{1 \: km/hr = \dfrac{5}{18} \: m/s }$

This implies ;

54 km/hr = ? m/s

=> 54 x 5/18

=> 3 x 5

=> 15 m/s

54 km/hr = 15 m/s

However,

0 km/hr = 0 m/s

Similarly,

1 minute = 60 seconds

5 minutes = ? seconds

=> 5 X 60

=> 300 seconds

Hence,

• Initial velocity ( u ) = 0 m/s

• Final velocity ( v ) = 15 m/s

• Time ( t ) = 300 seconds

Using the 1st equation ;

➱ v = u + at

➱ 15 = 0 + a x 300

➱ 15 = 300a

➱ 300a = 15

➱ a = 15/300

➱ a = 1/20

Hence,

★ Acceleration ( a ) = 1/20 m/s²

Now,

Using the 2nd equation of motion ;

↠ v² - u² = 2as

↠ ( 15 )² - ( 0 )² = 2 x 1/20 x s

↠ 225 - 0 = 2 x 1/20 x s

↠ 225 = 1/10 x s

↠ s/10 = 225

↠ s = 225 x 10

↠ s = 2250

Hence,

★ Displacement ( s ) = 2250 meters ( or 2.25 km )

Therefore,

☞ Acceleration = 1/20 m/s² ( or ) 0.05 m/s²

☞ The distance travelled to attain this velocity is 2250 meters ( 2.25 km )