Question

A train is travelling at a speed of 90km h ^-1 . Brakes are applied so 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:

625 m

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 90 km/h
• Acceleration (a) = -0.5 m/s²
• Final velocity (v) = 0 (As it comes to rest)

Converting initial velocity in m/s :

$\longrightarrow\sf{ u = 90\; kmh^{-1}}$

$\longrightarrow\sf{ u = \Bigg ( 90 \times \dfrac{5}{18} \Bigg ) \; ms^{-1}}$

$\longrightarrow\sf{ u = \Bigg ( 5 \times 5\Bigg ) \; ms^{-1}}$

$\longrightarrow\boxed{\sf{ u = 25 \; ms^{-1}}}$

Now, by using the third equation of motion :

$\longrightarrow$ v² - u² = 2as

• s denotes distance
• u denotes initial velocity
• a denotes acceleration
• v denotes final velocity

$\longrightarrow \sf {(0)^2 - (25)^2 = 2 \times (-0.5) \times s}$

$\longrightarrow \sf {0-625 = -1s}$

$\longrightarrow \sf {-625 = -1s}$

$\longrightarrow \sf {\cancel{\dfrac{-625 }{-1}}= s}$

$\longrightarrow \boxed{\sf {625 \; m = s}}$

Therefore, distance travelled before coming to rest is 625 m.

$\rule{200}2$

More Information:

First equation of motion :

$\longrightarrow$ v = u + at

Second equation of motion :

$\longrightarrow$ s = ut + ½at²

Third equation of motion :

$\longrightarrow$ v² = u² + 2as

Where,

• v denotes final velocity

• u denotes initial velocity

• a denotes acceleration

• s denotes distance

• t denotes time