Question

A train starting from rest moves with a
uniform acceleration of 0.2 m/s² for 5 min.
Calculate the speed acquired and the distance travelled in this time​

Answer 1

Answer :

➥ Final velocity of a train = 60 m/s

➥ Distance travelled by a train = 9 km

Given :

➤ Intial velocity of a train (u) = 0 m/s

➤ Acceleration of a train (a) = 0.2 m/s²

➤ Time taken by a train (t) = 5 min

To Find :

➤ Final velocity of a train (v) = ?

➤ Distance travelled by a train (s) = ?

Solution :

◈ Time taken (t) = 5 min = 5 × 60 = 300 sec

Final velocity of a train

From first equation of motion

$\tt{:\implies v = u + at}$

$\tt{:\implies v = 0 + 0.2 \times 300}$

$\tt{:\implies v = 0 + 60}$

$\bf{:\implies \underline{ \: \: \underline {\purple{ \: \: v = 60 \: m/s \: \: }} \: \: }}$

Hence, the speed required by the train is 60 m/s.

Distance travelled by a train

From second equation of motion

$\tt{:\implies s = ut + \dfrac{1}{2} {at}^{2} }$

$\tt{:\implies s = 0 \times 300 + \dfrac{1}{2} \times 0.2 \times 300 \times 300 }$

$\tt{:\implies s = 0 + 1 \times 0.2 \times 150 \times 300}$

$\tt{:\implies s = 0 + 0.2 \times 150 \times 300}$

$\tt{:\implies s = 0 + 30 \times 300}$

$\tt{:\implies s = 0 + 9000}$

$\bf{:\implies s = 9000 \: m}$

Convert into km :

$\tt{:\implies s = \cancel{\dfrac{9000}{1000} }}$

$\bf{:\implies \underline{ \: \: \underline {\purple{ \: \: s = 9 \: km \: \: }} \: \: }}$

Hence, the distance travelled by a train is 9 km.

⠀⠀⠀⠀

Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as

Answer 2

➭ Question :-

$⟶$ A train starting from rest moves with a uniform acceleration of 0.2 m/s² for 5 min. Calculate the speed acquired and the distance travelled in this time

➭ Given :-

$⟶$initial speed = 0.2 m/s²

$⟶$time taken = 5 min = 300 s

➭ To find :-

$⟶$the speed acquired and the distance travelled in this time

➭ Formula required :-

$⟶$$\boxed{\sf V = u + at}$

$⟶$$\boxed{\sf S = ut + 1/2at²}$

➭ Solution :-

let final speed = v

then, using kinematics equation

$⟹ v = u + at \: \: \: \: \: \: \: \\ ⟹ 0 + 0.2 \times 300 \: \: \\ ⟹ 60m/s \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

the speed of the train is 60m/s

$⟹distance \: travelled \: s = ut + \frac{1}{2} {at}^{2} \\ ⟹ 0 + \frac{1}{2} \times 0.2 \times (300)^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ⟹ 0.1 \times 90000 = 9000m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\by \: converting \: \: m \: into \: km \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ⟹ s = \frac{9000}{1000} = 9km \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

${\bold{\boxed{S = 9km }}}$

Hence, the speed acquired and the distance travelled in 300s is 9km