Question

(iii) A train starting from stationary position and
moving with uniform acceleration attains a speed of
36 km/h in 10 minutes. Find its acceleration.​

Answer 1

Answer:-

The Required Acceleration Of Train Is 0.016 m/s².

Explanation:-

Given:-

• Intial speed of train = 0m/s (As the train is at stationary position).

• Final Speed = 36km/h.

• Time = 10 minutes.

To Find:-

• Acceleration.

Formula used:-

$\\ \large\purple{ \boxed{ \red{1.} \: \: { \pink{ \bf a = \dfrac{v - u}{t}.}}}}$

Where,

• a = Acceleration.

• v = Final Velocity.

• u = Initial Velocity.

• t = Time.

Now,

Lets convert all these units in the same unit.

▪︎ Initial Velocity = 0 m/s = u.

▪︎ Final Velocity $\tt = 36km/h = \dfrac {36\times1000}{1×3600} m/s = 10m/s = v.$

▪︎ Time $\tt= 10\:min = 10\times60\:sec = 600\:sec= \:t.$

So,

By Using formula (1), Acceleration by train,

$\\ \bf = \dfrac{v-u}{t}.$

$\\ \tt = \dfrac{10m/s - 0m/s}{600s}.$

$\\ = \dfrac{10m/s}{600s}.$

$\\ =\dfrac{1}{60} m/s^2.$

$\\ \large\red{\boxed{\blue{\bf = 0.016\: m/s^2.}}}\ \ \ \bf(Approx)$

Therefore Acceleration Of the train is equal to 0.016 m/s².