Question

A train accelerated from 20km/hr to 80km/hr in 4minutes how much distance does it cover in the period? Answer that the tracks are straight

Answer 1

Answer:

3,354.8 metres

Explanation:

Given :

• Initial velocity = u = 20 km/hr = 20×5/18 = 50/9 m/s

• Final velocity = v = 80 km/hr = 80×5/18 = 200/9 m/s

• Time taken = t = 4 minutes = 240 seconds

To find :

• Distance travelled by the train in this period

Using the first equation of motion :

V=u+at

200/9 = 50/9+a×240

150/9 = 240a

a = 150/9 × 1/240

a = 0.069 m/s²

Using the third equation of motion :

V²-u²=2as

(200/9)²-(50/9)²=2×0.069×s

(37,500/81)=0.138s

462.96 = 0.138s

S = 3,354. 8 metres

The distance covered by the train is equal to 3,354.8 metres

Answer 2

Answer:

Explanation:

Solution,

Here, we have

Initial velocity, u = 20 km/h = 20 × 5/18 = 5.55 m/s

Final velocity, v = 80 km/h = 80 × 5/18 = 22.22 m/s

Time taken, t = 4 minutes = 4 × 60 = 240 seconds

To Find,

Distance covered, s = ??

At first we have to find acceleration, a

According to the 1st equation of motion,

We know that

v = u + at

So, putting all the values, we get

⇒ v = u + at

⇒ 22.22 = 5.55 + a × 240

⇒ 22.22 - 5.55 = 240a

⇒ 16.67 = 240a

⇒ 16.67/240 = a

⇒ a = 0.069 m/s²

Here, the acceleration is 0.069 m/s²

According to the 3rd equation of motion,

We know that

v² - u² = 2as

So, putting all the values, we get

⇒ v² - u² = 2as

⇒ (22.22)² - (5.55)² = 2 × 0.069 × s

⇒ 493.72 - 30.8 = 0.138s

⇒ 462.92 = 0.138s

⇒ 462.92/0.138 = s

⇒ s = 3354.49 m.

Hence, the distance covered by train is 3354.49 m.