Question

a train starting from a Railway Station and moving with uniform acceleration attains a speed of 40km/h in 10mins it's acceleration is....1)18.5m/s2. 2)1.85m/s2. 3)18.5cm/s2. 4)1.85cm/s2​

Answer 1

Given:-

• Initial velocity ,u = 0m/s
• Final velocity ,v = 40km/h = 40×5/18 = 11.11m/s
• Time taken ,t = 10min = 10×60 = 600s

To Find:-

• Acceleration ,a

Solution:-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀According to the Question

It is given that the train start from rest & attain velocity of 40km/h in 10 min. We have to calculate the acceleration of the train .

Acceleration is defined as the rate of change of velocity at per unit time.

• a = v-u/t

where,

• v is the final velocity
• a is the acceleration
• u is the initial velocity
• t is the time taken

Substitute the value we get

$:\implies$ a = 11.11-0/600

$:\implies$ a = 11.11/600

$:\implies$ a = 0.0185 m/s²

now , converting the acceleration into cm/s² . so we multiply with 100

$:\implies$ a = 0.0185 × 100 cm/s²

$:\implies$ a = 1.85cm/s²

• Hence, the acceleration of the train is 1.85cm/s².

So , the required option is (4) 1.85cm/s².

Answer 2

Question:-

• A train starting from a Railway Station and moving with uniform acceleration attains a speed of 40km/h in 10mins it's acceleration is

To Find:-

• Find the acceleration.

Solution:-

Here ,

• initial velocity = 0

• final velocity = 11.11 m/s

• time = 600 seconds

First ,

We have to find the acceleration:-

$\dashrightarrow\sf \: v = u + at$

$\dashrightarrow\sf \: 11.11 = 0 + 600a$

$\dashrightarrow\sf \: a = \cancel\dfrac { 11.11 } { 600 }$

$\dashrightarrow\sf \: a = 0.0185 \: { m/s }^{ 2 }$

Now ,

We have to change m/s to cm/s:-

$\dashrightarrow\sf \: 0.0185 \times 100$

$\dashrightarrow\sf \: 1.85 \: cm/s$

Hence ,

• Acceleration is 1.85 cm/s