Question

A train starting from rest attains a velocity of 72 km per hr in 5 minutes. Assuming that the acceleration is uniform. The distance travelled by the train for attaining this velocity will be

Answer 1

Given :

• initial velocity of train, u = 0
• final velocity of train, v = 72 km h⁻¹
• time taken for changing velocity , t = 5 min = 5/60 hrs = 1/12 hrs

To find :

• Distance travelled by the train for attaining velocity 72 km h⁻¹ , s = ?

Formulae used :

• First equation of motion

      v = u + a t

• second equation of motion

     s = u t + 1/2 a t²

( where v is final velocity , u is initial velocity , a is acceleration , t is time and s is distance covered )

Calculation :

Calculating acceleration of train , a

Using first equation of motion

→ v = u + a t

→ 72 = 0 + a ( 1/12 )

→ a = 72 × 12

→ a = 864  km h⁻²

Calculating distance covered by train to attain velocity 72 km h⁻¹ , s

Using second equation of motion

→ s = u t + 1/2 a t²

→ s = (0) (1/12) + 1/2 (864) (1/12)²

→ s = 1/2 (864) (1/144)

→ s = 3 km

therefore,

▶Distance travelled by train for attaining final velocity is 3 km .

Answer 2

$\sf\large{\underline{Solution :-}}$

• Initial velocity (u) = 0
• final velocity (v) = 72 km/h = 20 m/s and
• time (t) 5 minute 300 s

◆ Equation (¡) = a = (v-u)/t

→ ( 20 m/s-0 m/s )/ 300s

→ 1 / 15 m/s

→ 2as = v² - u² = v² -0

→ S = v² /2a

→ ( 20 m/s )²/ 2x (1/15) m/s

→ 400 / 2 × 1/15 m/s

→ 200 × 1/15 m/s

→ 200 × 15 m/s

→ 3000 m

→ 3000/1000 km. (changing metre in km )

→ 3 km

$\sf\large{\underline{Hence :-}}$

★ The acceleration of the train is 1/15 m/s

★ And the distance travelled is 3 km.