Question

A train starting from a railway station and moving with inform acceleration, attains a speed of 40 km/h in 10 minutes, Is acceleration is :

Answer 1

Given:-

• Train starting from station
• Train attain speed of 40km/h
• Time taken to attain speed = 10 minutes.

To Find:

• Acceleration?

Solution:-

If train is starting from station that mean in present, train is at rest therefore, initial velocity (u) is 0 and final velocity (v) is 40km/h. Time interval is 10 minutes

Making same unit :-

• Initial velocity = 0m/s
• Final velocity = 40km/h = 40×5/18 = 200/18 = 11.11m/s
• Time = 10 minutes = 10×60 = 600seconds

We know that :

• $\large{\boxed{\boxed{\small{\mathfrak{\red{Acceleration~=~\dfrac{v-u}{t}}}}}}}$

Therefore,

$\small{\tt{Acceleration~=~\dfrac{11.11-0}{600}}}$

$\implies\small{\tt{Acceleration~=~\dfrac{11.11}{600}}}$

$\implies\small{\tt{Acceleration~=~0.0185m/s^2}}$

Answer 2

Answer:

Given:-

Train starting from station

Train attain speed of 40km/h

Time taken to attain speed = 10 minutes.

To Find:

Acceleration?

Solution:-

If train is starting from station that mean in present, train is at rest therefore, initial velocity (u) is 0 and final velocity (v) is 40km/h. Time interval is 10 minutes

Making same unit :-

Initial velocity = 0m/s

Final velocity = 40km/h = 40×5/18 = 200/18 = 11.11m/s

Time = 10 minutes = 10×60 = 600seconds

We know that :

\large{\boxed{\boxed{\small{\mathfrak{\red{Acceleration~=~\dfrac{v-u}{t}}}}}}}

Acceleration =

t

v−u

Therefore,

\small{\tt{Acceleration~=~\dfrac{11.11-0}{600}}}Acceleration =

600

11.11−0

\implies\small{\tt{Acceleration~=~\dfrac{11.11}{600}}}⟹Acceleration =

600

11.11

\implies\small{\tt{Acceleration~=~0.0185m/s^2}}⟹Acceleration = 0.0185m/s

2