Question

A train is travelling at a speed of 75 km/h. The brakes are applied so as to produce a uniform acceleration of -0.5 m/s2 . Find how far the train goes before it stops?

Answer 1

GiveN :

• Initial velocity (u) = 75 km/h
• Final velocity (v) = 0 m/s
• Acceleration (a) = - 0.5 m/s²

To FinD :

• How far train goes before it stops

SolutioN :

We are given that train was travelling with speed of 75 km/h and breaks are applied, it means train well stop after applying brakes. and final velocity (v) is 0 m/s.

Convert Initial velocity into m/s.

⇒Initial velocity (u) = 75 * 5/18 = 20.83 m/s (approx).

Use 3rd equation of motion :

⇒v² - u² = 2as

⇒0² - (20.83)² = 2 * - 0.5 * s

⇒0 - 433.89 = -s

⇒- s = - 433.89

⇒s = 433.89

∴ Distance travelled by train after applying breaks is 433.89 m (approx.)

Answer 2

$\huge{\underline{\sf{\bigstar{ANSWER}}}}\bigstar$

$\bigstar\sf\blue{GIVEN}\bigstar$

• $\bigstar{\sf{\blue{INITIAL\:VELOCITY{u}=75km/hr=20.8m/s^{-1}}}}\bigstar$

• $\bigstar\sf\red{Final\:velocity=0}\bigstar$

• $\bigstar\sf\pink{Acceleration=-0.5m/s^-2}\bigstar$

$\bigstar\sf\blue{TO\:FIND}\bigstar$

• Distance travelled by train.

$\bigstar\sf\blue{HOW\:TO\:SOLVE}\bigstar$

• We have given that train aplies brakes so the trains stops so the final velocity will be 0.

Now,

Using third equation of motion i.e

$\bigstar\tt{\boxed{V^2=U^2-2as}}\bigstar$

Put the values.

$\implies\sf{(o)^2=(20.8)^2-2×0.5×s}$

$\implies\sf{-432.64=-2×0.5s}$

$\implies\sf{s=431.64m}$

Hence, The distance travelled by train after applying brakes is 431.64m approx.