Question

The brakes are applied to a train produce an acceleration of 5m/s.sq in the opposite direction to the motion.if the train produce 6 second to stop after applying brake calculate distace travelled duri g this time

Answer 1

Given :

➾ Acceleration of train due to applied brakes = 5m/s²

➾ Time taken by train to stop = 6s

To Find :

➨ Distance travelled by train before it is brought to rest.

Concept :

➠ This question is completely based on the concept of stopping distance.

➠ In this question, First we have to find out initial velocity of train after that we can calculate distance travelled by train.

➠ Since, Acceleration has said to be constant throughout the period of journey, We can easily apply equation of kinematics to solve this type of questions.

➠ Train comes to rest after applying retarding force hence final velocity of train will be zero.

Calculation :

✭ Initial velocity of train :

➝ v = u + at

➝ 0 = u + (-5 × 6)

➝ 0 = u + (-30)

➝ u = 30 m/s

✭ Distance covered by train :

⇒ v² - u² = 2as

⇒ (0)² - (30)² = 2(-5)s

⇒ -900 = -10s

⇒ s = 900/10

⇒ s = 90m

[Note : -ve sign of a shows retardation.]

Answer 2

$\huge\sf\pink{Answer}$

☞ Distance = 90 m

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$\huge\sf\blue{Given}$

✭ Acceleration (a) = - 5 m/s

✭ Time (t) = 6 second

✭ Final Velocity (v) = 0 m/s

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$\huge\sf\gray{To \:Find}$

◈ Distance Travelled?

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$\huge\sf\purple{Steps}$

So here we should first find the initial velocity of the body,we can find it with the help of,

$\underline{\boxed{\sf v = u+at}}$

Substituting the given values,

➳ $\sf 0 = u+(-5)(6)$

➳ $\sf -u = -30$

➳ $\sf\red{u = 30 \ m/s}$

Now we shall find the distance travelled by using the second Equation of motion,that is,

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

Substituting the values,

»» $\sf 0^2-30^2 = 2 \times (-5) \times s$

»» $\sf -900 = -10s$

»» $\sf\dfrac{-900}{-10} = s$

»» $\sf\orange{s = 90 \ m}$

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