Question

A train running with a velocity of 10m/s attains an acceleration of 2m/s . What is the distance covered by it in 10 sec?

Answer 1

Answer:

An object is moving with an initial velocity 10m/s(-2minute per second square)then total distance covered by the object in 6 seconds

Explanation:

Answer 2

Answer:

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)