Question

Speed of a train increases from 36km/h to 72km/h in 10 minutes. Find distance
covered by the train during this time.

Answer 1

Answer:

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Speed of a train increases from 36km/h to 72km/h in 10 minutes. Find distance covered by the train during this time.

Correct Question :

❤✔Speed of a train increases from 36km/h to 72km/h in 10 seconds Find distance covered by the train during this time.

Solution :

u = 36km/h = 10m/s

v = 72km/h = 20m/s

t = 10sec =

a = v -u/t

a = 20-10/10

a = 1m/s^2

s = UT +1/2at^2

s = 10*10 + 1/2*1 * 100

s = 100 + 50

s = 150m

___ΔVASU ❤____________@ZQUAD

Answer 2

$\huge{\mathfrak{\red{\underline{\underline{........Correct\;Question........}}}}}$

Speed of a train increases from 36km/h to 72km/h in 10 sec. Find distance covered by the train during this time.

$\huge{\mathfrak{\red{\underline{\underline{........Solution........}}}}}$

Given:

=> Initial Velocity (u) = 36 km/h

                            = 36 × 5/18 m/s

                            = 10 m/s.

=> Final Velocity (v) = 72 km/h

                           = 72 × 5/18 m/s

                           = 20 m/s.

=> Time (t) = 10 sec

To Find:

=> Distance covered by train.

Formula used:

$\sf{\implies a = \dfrac{v-u}{t}}$

$\sf{\implies s = ut+\dfrac{1}{2}at^{2}\;\;\;\;[2nd\;equation\;of\;motion]}$

So, first we will find acceleration,

$\sf{\implies a = \dfrac{v-u}{t}}$

$\sf{\implies a = \dfrac{20-10}{10}}$

$\sf{\implies a = \dfrac{10}{10}}$

$\large{\boxed{\boxed{\purple{\sf{\implies a = 1\;m/s^{2}}}}}}$

Now, we will find distance by 2nd Equation of motion.

$\sf{\implies S = ut+\dfrac{1}{2}at^{2}}$

Put the values in the formula, we get

$\sf{\implies s = 10\times 10+\dfrac{1}{2}\times 1\times 10^{2}}$

$\sf{\implies s = 100 + \dfrac{1}{2} \times 100}$

$\sf{\implies s = 100 + 50}$

$\large{\boxed{\boxed{\green{\sf{\implies s = 150\;m}}}}}$

Hence, Distance travelled by train is 150 m.