Question

A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h−1 in 10 minutes. Find its acceleration. 0.172 m/s² 0.185 m/s² 0.956 m/s² None of these please answer this​

Answer 1

Answer:

The required acceleration is 0.0185 m/s²

Explanation:

Given :

A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/h in 10 minutes.

To find :

the acceleration

Solution :

Initially, the train is at rest.

So, initial speed, u = 0 m/s

Final speed of the train, v = 40 km/h

Converting km/h to m/s , we get :

1 km/h = 5/18 m/s

40 km/h = 40 × 5/18 m/s

v = 200/18 m/s

v = 100/9 m/s

v = 11.11 m/s

Time taken, t = 10 min = 10 × 60 s = 600 s

We know that the acceleration is the rate of change of velocity.

$\longmapsto \tt a=\dfrac{v-u}{t}$

Substituting the values,

$\sf a=\dfrac{11.11-0}{600} \\\\ \sf a=\dfrac{11.11}{600} \\\\ \sf a=0.0185 \ m/s^2$

Therefore, the required acceleration is 0.0185 m/s²

Answer 2

Answer:

Answer:

The required acceleration is 0.0185 m/s²

Explanation:

Given :

A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/h in 10 minutes.

To find :

the acceleration

Solution :

Initially, the train is at rest.

So, initial speed, u = 0 m/s

Final speed of the train, v = 40 km/h

Converting km/h to m/s , we get :

1 km/h = 5/18 m/s

40 km/h = 40 × 5/18 m/s

v = 200/18 m/s

v = 100/9 m/s

v = 11.11 m/s

Time taken, t = 10 min = 10 × 60 s = 600 s

We know that the acceleration is the rate of change of velocity.

\longmapsto \tt a=\dfrac{v-u}{t}⟼a=

t

v−u

Substituting the values,

\begin{gathered}\sf a=\dfrac{11.11-0}{600} \\\\ \sf a=\dfrac{11.11}{600} \\\\ \sf a=0.0185 \ m/s^2\end{gathered}

a=

600

11.11−0

a=

600

11.11

a=0.0185 m/s

2

Therefore, the required acceleration is 0.0185 m/s²