Question

A train is moving with a velocity of 10m/s. It attains an acceleration of 4m/s after 5 seconds. Find the distance covered by the train at that time.​

Answer 1

Given that:

• A train is moving with a velocity of 10 m/s.
• It attains an acceleration of 4 m/s² after 5 seconds.

To Find:

• The distance covered by the train at that time.

First equation of motion,

• v = u + at

Where,

• v = Final velocity
• u = Initial velocity = 10 m/s
• a = Acceleration = 4 m/s²
• t = Time = 5 seconds

Finding the final velocity:

⟶ v = 10 + (4 × 5)

⟶ v = 10 + 20

⟶ v = 30

∴ Final velocity = 30 m/s

Second equation of motion,

• s = ut + (at²)/2

Where,

• s = Displacement
• v = Final velocity = 30 m/s
• u = Initial velocity = 10 m/s
• a = Acceleration = 4 m/s²
• t = Time = 5 seconds

Finding the distance covered by the train:

⟶ s = (10 × 5) + (4 × 5²)/2

⟶ s = 50 + (2 × 25)

⟶ s = 50 + 50

⟶ s = 100

Hence,

• The distance covered by the train at that time is 100 m.

Answer 2

Answer:

Given :-

• A train is moving with a velocity of 10 m/s.
• It attains an acceleration of 4 m/s² after 5 seconds.

To Find :-

• What is the distance covered by the train at that time.

Formula Used :-

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2} at^2}}}$

where,

• u = Initial Velocity
• a = Acceleration
• t = Time
• s = Distance Covered

Solution :-

Given :

• Initial Velocity (u) = 10 m/s
• Time (t) = 5 seconds
• Acceleration (a) = 4 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf s =\: (10)(5) + \dfrac{1}{2} \times (4)(5)^2$

$\longrightarrow \sf s =\: 10 \times 5 + \dfrac{1}{2} \times 4 \times 5 \times 5$

$\longrightarrow \sf s =\: 50 + \dfrac{1}{2} \times 20 \times 5$

$\longrightarrow \sf s =\: 50 + \dfrac{1}{\cancel{2}} \times {\cancel{100}}$

$\longrightarrow \sf s =\: 50 + 50$

$\longrightarrow \sf\bold{\red{s =\: 100\: m}}$

$\therefore$ The distance covered by the train is 100 m .

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EXTRA INFORMATION :-

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ Third Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}$

where,

• s = Distance Covered (m)
• u = Initial Velocity (m/s)
• v = Final Velocity (m/s)
• a = Acceleration (m/s²)
• t = Time (s)