A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform. The distance travelled by train for the above velocity is

CONCEPT:

• Equation of Kinematics: These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
• Equations of motion can be written as

V = U + at

\(s =ut+\frac{1}{2}{at^{2}}\)

V2 =U2+ 2as

Where, U = Initial velocity, V = Final velocity, g= Acceleration due to gravity, t = time, and h= height/Distance covered

Where u = Initial velocity of the particle at time t = 0 sec

 v = Final velocity at time t sec

a = Acceleration of the particle

s = Distance travelled in time t sec

EXPLANATION:​

Given - Initial velocity (u) = 0 km/h, Final velocity (v) = 72 km/h = 20 m/s  and Time (t) = 5 min.

• As per the first law of motion

⇒ v = u + a t

⇒ 20 = 0 + (a₁ × 5 × 60)

\(⇒ a = \frac{{20}}{{5 \times 60}} = \frac{1}{{15}}\, ms^2\)

• The distance travelled by train 

\(⇒ S = u t + \left( {\frac{1}{2} \times a \times {t^2}} \right)\)

\(⇒ S = 0 \times 300 + \left( {\frac{1}{2} \times \frac{1}{{15}} \times {{300}^2}} \right)\)

⇒ S = 3000 m = 3 km

• The distance travelled by train for attaining the above velocity is 3 km.