Question

A train is traveling at a speed of 80 km per hour. Breaks are applied as to produce a uniform acceleration of -0.5ms^-2. Find how far the train will go before it is brought to rest​

Answer 1

Answer:

$\bigstar{\bold{Distance\:travelled=493.73\:m}}$

Explanation:

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

• Initial velocity (u) = 80 km/hr = 22.22 m/s
• Final velocity (v) = 0 m/s
• Acceleration (a) = -0.5 m/s²

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

• Distance travelled(s)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Here we have to find the distance travelled by the train before it is brought to rest.

→ By the third equation of motion, we know that

  v² - u² = 2as

→ Substituting the given datas,

  0² - 22.22² = 2 × -0.5 × s

 - 493.73 = -1 s

   s = 493.73 m

→ Hence the distance travelled by the train is 493.73 m

$\boxed{\bold{Distance\:travelled=493.73\:m}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as

Answer 2

Answer:

Distance travelled = 493.73m