Question

a train starting from rest attains a velocity of 72 km per hour in 5 minute as human the acceleration is uniform find
1) the acceleration
2) the distance travelled by the train for attaining this velocity ​

Answer 1

Answer:

The acceleration is 1/15 m/s² or 0.066m/s²

and the distance travelled is 3000m or 3km

$\large \bf \underline{Given :}$

• A train starting from rest attains a velocity of 72km/h in 5 minutes with uniform acceleration

$\large \bf\underline{To \: Find : }$

• The acceleration
• The distance travelled by the train to attain the velocity

$\bf\large\underline{Solution}$

Given,

Initial velocity , u = 0m/s

Final velocity , v = 72km/h

$\sf\longrightarrow v = 72\times \dfrac{5}{18} ms^{-1} \\\\ \sf\implies v = 4\times 5 ms^{-1} \\\\ \sf\implies v = 20ms^{-1}$

time taken ,

t = 5 mins

$\sf\implies t = 5\times 60 s \\\\ \sf\implies t = 300 s$

Now using the equations of motion :

$\sf\implies v = u + at \\\\ \sf\implies a = \dfrac{v-u}{t} \\\\ \sf\implies a = \dfrac{20 - 0}{300} \\\\ \sf\implies a = \dfrac{20}{300} \\\\ \sf\implies a = \dfrac{1}{15} \\\\ \sf \implies a = 0.066$

Therefore,

acceleration is 0.066 ms²

Again, from equations of motion

$\sf\implies s = ut+\dfrac{1}{2}at^{2} \\\\ \sf\implies s = 0\times 300 + \dfrac{1}{2} \times \dfrac{1}{15}\times 300\times 300 \\\\ \sf\implies s = 0 + 10\times 300 \\\\ \sf\implies s = 3000$

Thus , the distance travelled is 3000m or 3km

Answer 2

✤ Correct Question :-

A train starting from rest attains a velocity of 72 km per hour in 5 minute assuming that acceleration is uniform . find

1) the acceleration

2) the distance travelled by the train for attaining this velocity

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✤ Answer

⠀⠀⠀⠀⠀⠀1) Acceleration = 864 km/hr²

⠀⠀⠀⠀⠀⠀2) Distance ⠀⠀= 3km

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✤ Explanation :-

1) Acceleration

${\sf{\underline{\underline{\green{\sf{ Given\:, }}}}}}$

Initial Velocity ( u ) = 0

final Velocity ( v ) = 72 km/hr

Time taken ⠀ ( t ) = 5 min

➢( we have to change minutes into hours )

⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ 1 hr = 60min

⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ 5minutes = hours?

⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ➪ 5/60 = 1/12 hrs

Acceleration (a) = ?

⠀ ⠀ ⠀ ⠀ ⠀ ⠀ We know that , a = v - u / t

$acceleration \: = \frac{final \: velocity - initial \: velocity}{time}$

$a \: = \frac{72 - 0}{ \binom1{12} }$

$a \: = \: \frac{72}{ \binom{1}{12}}$

$a \: = \: \frac{72 \times 12}{1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (72 \div \frac{1}{12} )$

$a \: = \: 864$

a = 864 km/hr²

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2) the distance travelled by the train for attaining this velocity

${\sf{\underline{\underline{\green{\sf{ Given\:, }}}}}}$

Initial Velocity ( v ) = 0

final Velocity ( u ) = 72 km/hr

Time taken ⠀ ( t ) = 5 min

Distance travelled ( s ) = ?

⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ ⠀ We know that , s = ut + ½ at²

( by using second equation of motion )

$s = ut + \frac{1}{2}{at}^{2}$

$s = 0 \times \frac{1}{12} + \frac{1}{2} \times 864 \times ( \frac{1} {12})^{2}$

$s = 1 \times \frac{1}{2} \times 864 \times \frac{1}{144}$

$s = 1 \times 432 \times \frac{1}{144}$

$s = 432 \times \frac{1}{144}$

$s = 3$

Distance ( s ) ➾ 3 km

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${\sf{\underline{\boxed{\red{\sf{ hence , }}}}}}$

✒${\sf{\underline{\orange{\sf{ Acceleration\:=\:864km/hr²}}}}}$

✒${\sf{\underline{\red{\sf{ Distance \: = \:3km}}}}}$

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☞ Additional information :-

➸ ✔ The three equations of motion that describe the behavior of a physical system in terms of position, Velocity and acceleration as functions of time are :

i) v = u + at

ii) s = ut + ½at²

iii) v² - u² = 2as

➸ ✔ These equations of motion are valid only when the acceleration is constant .

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