Question

$\huge\mathfrak\red{Question}$

If the velocity of a train which starts from rest is 72 km/h after 5 minutes , find out its acceleration and the distance travelled by the train in this time .

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$\small\mathrm\green{✷Need\:step\: by\: step\: explanation}$

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Answer 1

Answer:

14.40 is the correct answer

Answer 2

Given:

• Initial velocity (u) = 0
• Final velocity (v) = 72km/h => $\large{\bf{72×\frac{5}{18}=20m/s}}$
• Time = 5minute = 5×60 = 300 second.

To find:

• Acceleration of train
• Distance travelled by train in given time.

Solution:

Finding acceleration of train :-

We are given that initial velocity and final velocity of train are 0 and 20m/s respectively in time 300 second.

We know that:

$\large{\tt{\green{Acceleration (a)\:=\: \frac{final\: velocity (v)-initial \: velocity (u)}{time\: taken (t)}}}}$

Putting known values in formula :-

Acceleration = $\large{\sf{\dfrac{v-u}{t}}}$

→ Acceleration = $\large{\sf{\dfrac{20 -0}{300}}}$

→ Acceleration = $\large{\sf{\dfrac{20}{300}}}$

→ Acceleration = $\large{\sf{\cancel{\dfrac{20}{300}}}}$

→ Acceleration = $\large{\sf{\dfrac{1}{15}}}$

Therefore,

$\huge{\boxed{\large{\pink{\tt{Acceleration\: of \: train\:=\: {\dfrac{1}{15}m/s²}}}}}}$

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Finding distance travelled by train :-

We have some given values are as :-

• Initial velocity (u) = 0
• Final velocity (v) = 20m/s
• Time(t) = 300 second.
• Acceleration (a) = $\large{\bf{\dfrac{1}{15}m/s²}}$

We know that :

According the the Newton's second laws of motion,

$\large{\boxed{\tt{\green{s\: =\: ut+\frac{1}{2} at²}}}}$

Where:

• s = distance
• u = initial velocity
• t = time
• a = acceleration

Putting known values in formula:

$\large{\sf{s\: =\: ut+\frac{1}{2} at²}}$

→$\large{\sf{s\: =\:0×300+\frac{1}{2}×\frac{1}{15}×300²}}$

→ $\large{\sf{s\: =\: \frac{1}{30}×90000}}$

→ $\large{\sf{s \: =\: {\frac{1}{\cancel{30}}}} \times\cancel {90000}}$

→ $\large{\sf{s\:=\: 3000m}}$

Therefore,

$\huge{\boxed{\large{\tt{\green{Distance\: travelled\: by\: train \: is\: 3000m\: or\: 3km}}}}}$