Question

A train is travelling at a speed of 90km/hr. The brakes were applied so as to produce a uniform acceleration gf - 0.5 m/s ^ 2 . Find how far the train goes before it stops. SES​

Answer 1

Solution :-

Given ,

• Speed of train = 90km/h = 25 m/s
• Acceleration of the car after applying brakes = - 0.5 m/s²

We need to find ,

• Distance travelled by the train after applying brakes .

Finding distance travelled after applying brakes using the kinematic equation,

♦ v² - u² = 2as

Where ,

• v → final velocity = 0 m/s
• u → initial velocity = 90 km/h = 25 m/s
• a → acceleration = - 0.5 m/s²
• s → distance = ?

• Note :- Here , the final velocity of the train will be zero because the train is stopped by applying brakes so the final velocity will be zero .

Substituting known values we have,

⇒ 0² - 25² = 2 × ( - 1/2 ) × ( s )

⇒ 0 - 625 = - s

⇒ - 625 = - s

⇒ Distance travelled after applying brakes = 625 m = 0.625 km

Hence , distance travelled after applying brakes = 0.625 km = 625 m

Answer 2

Answer:

625 m

Explanation:

Given:

• Initial velocity (u) = 90 km/h = 90 × 5/18 = 25 m/s
• Final velocity (v) = 0 m/s
• Acceleration (a) = -0.5 m/s^2

To find:

How far the train will go before it stops or distance (s) = ?

Solution:

We know Newton's third kinematic equation i.e, v^2 - u^2 = 2as

(putting values)...

=> (0)^2 - (25)^2 = 2 × -0.5 × s

=> 0 - 625= -1 × s

=> -625 = -1 × s

=> - 625/- 1 = s

=> 625 m = s

Distance = 625 meters

Hence, we get the required ans

Hope this helped you dear...