Question

A train is travelling at a speed of 90km/hr brakes are applied to produce uniform acceleration -0.5m/s square find how far will the train go before it is brought to rest

Answer 1

hope you will satisfied with answer! Thankyou

Answer 2

We have given :

• initial velocity of train(u) = 90km/h

Converting in m/s ➟

$\small{\bf{90 \times \frac{5}{18} }}$

=> 25m/s

• Final velocity of the train (v)= 0 ( train come to rest )
• uniform Acceleration(a) = -0.5m/s

We have to find :

• Distance covered by train before come to rest

Solution :

In the question we have Speed of train and acceleration of train, and we have to find distance traveled by train.

Hence, we will use Third laws of motion.

Formula using :

Third laws of motion :

$\large{\sf{ {v}^{2} - {u}^{2} = 2as}}$

Where :

• v = Final velocity
• u = initial velocity
• a = acceleration
• s = distance

Let's substitute values in formula :

$\small{\sf{ {v}^{2} - {u}^{2} = 2as}}$

➟ $\small{\sf{ {0}^{2} - {25}^{2} =2 \times 0.5 \times s }}$

➟$\small{\sf{-625=\: -1×s}}$

➟ $\small{\sf{s=\: \frac{ - 625}{ - 1} }}$

➟$\large{\sf{\boxed{\green{s=625m}}}}$

Or,

$\large{\sf{\boxed{\green{Distance~ covered=625m}}}}$

Some other information :-

• To change km/h into m/s always multiply the given value with 5/18.

• Acceleration : Accelerationof a body is change in velocity of the body per unit time
• Distance : The actual length of path covered by a moving body between its initial and final position is called the distance covered by body
• Velocity : Velocity of a body is its displacement per unit time or speed in a definite direction

Three Equations of motion :

1. v = u + at² -----(in this equation, s is not present)
2. s = ut +at² -----( in this equation, v is not present)
3. v² - u² = 2as -----( in this equation, t is not present)

Where :

• v = final velocity
• u = initial velocity
• a = acceleration
• t = time
• s = distance