Question

A train starting from rest attains a speed of 25 m/s what will be distance after 20 sec

Answer 1

intial velocity, u = 0 m/s

final velocity, v = 25m/s

time, t = 20 sec

acceleration,

a = ( v – u ) / t

a = ( 25 – 0 ) / 20

a = 25 / 20

a = 5 /4 m/s²

Now using 2nd eq of motion

s = ut + 1/2at²

s = 0 + 1/2 ( 5/4 )( 20 )²

s = ( 5/8 )( 400 )

s = ( 5 × 50 )

s = 250 m

Ans : Distance travelled in 20 sec is 250 m

Answer 2

Given :-

A train starting from rest attains a speed of

25 m/s .

Required to find :-

• Distance covered after 20 seconds

Equations used :-

$\large{ \dagger \text{ \pink{v = u + at}}}$

$\large \dagger \pink{ \tt{{v}^{2} - {u}^{2}} = 2as }$

Solution :-

Given information :-

A train starting from rest attains a speed of

25 m/s

we need to find the distance covered after 20 seconds

From the given information we can conclude that ;

• Initial velocity of the train ( u ) = 0 m/s

• Final velocity of the train ( v ) = 25 m/s

• Time ( t ) = 20 seconds

Using the 1st equation ;

☛ v = u + at

☛ 25 = 0 + a x 20

☛ 25 = 20a

This implies ,

☛ 20a = 25

☛ a = 25/20

☛ a = 5/4 m/s²

Hence,

Acceleration of the car after 20 seconds = 5/4 m/s²

Using the next equation of motion ;

☛ v² - u² = 2as

☛ ( 25 )² - ( 0 )² = 2 ( 5/4 ) ( s )

☛ 625 - 0 = 5/2 x s

☛ 625 = 5/2 x s

☛ 625 x 2 /5 = s

This implies ,

☛s = 625 x 2/5

☛ s = 1250/5

☛ s = 250 meters

Therefore ,

Distance travelled after 20 seconds = 250 meters