Question

a train is travelling at abspeed of 90km per hr brakes are applied so as to produce a uniform acceleration of -o.5 m per second find hoew for the train will go before it brought to rest​

Answer 1

Answer :

• Distance covered by train after applying brakes and before coming to rest is 625 m

Explanation :

Given,

• initial velocity of train, u = 90 km/h

• uniform acceleration produced on applying brakes, a = -0.5 m/s²
• final velocity of train, v = 0  [ since it is brought to rest ]

To find,

• Distance covered by train before it is brought to rest after applying brakes, s = ?

Solution,

Converting initial velocity of train given in km/h into m/s

→ u = 90 km/h

→ u = 90 × ( 5 / 18 )  m/s

→ u = 25 m/s

Now,

Calculating distance covered by train after applying brakes and before coming to rest , s = ?

Using third equation of motion

→ 2 a s = v² - u²

→ 2 ( -0.5 ) ( s ) = ( 0 )² - ( 25 )²

→ - s = - 625

→ s = 625 m

therefore,

• Distance covered by train after applying brakes and before coming to rest is 625 metres .

Answer 2

${ \bold{ \text{ \underline{ Correct \: Question}}}}$

A train is travelling at speed of 90km/hr

brakes are applied so as to produce

a uniform acceleration of -0.50/sec square Find How far the train will go before it is brought to rest?

${ \bold{ \text{ \underline{Notations \: used:-}}}}$

u = initial velocity

a = acceleration

v = final velocity

s = distance travelled

${ \bold{ \text{ \underline{ Given:-}}}}$

• u = 90km/sec

$= 90 \times \dfrac{5}{18}$

${ \bf{ = 25m \: per \: sec}}$

• a = - 0.5m/s^2

Train will be is on rest so

• v= 0

${ \bold{ \text{ \underline{Solution:-}}}}$

${ \mathtt{ \red{ \implies \: v = u + at}}}$

${ \bf{ \implies0 = 25 - 2.5 \times t}}$

${ \bf{ \implies25 = 0.5 \times t}}$

${ \boxed{ \bf{ \implies t= 50sec}}}$

${ \mathtt{ \red{ \implies \: s = ut + \frac{1}{2} {at}^{2}}}}$

${ \bf{ \implies \: s = 25 \times 50 - \frac{1}{2} \times 0.5 \times {50}^{2}}}$

${ \bf{ \implies \: s = 1250 - \frac{1}{2} \times 0.5 \times 2500}}$

${ \boxed{ \bf{ \implies \: s = 625m}}}$

So Distance travelled by the train is 625m in time 50 sec

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