Question

A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find the acceleration and the distance travelled by train for attaining this velocity.

• A. 3 m/s² , 1/15 km
• B. 2 m/s² , 15 km
• C. 1/15 m/s² , 3 km
• D. 4.5 m/s² , 2 km

Answer 1

Answer: (C)

Step 1: Given data

The initial velocity of the train, u = 0

Final velocity of train, v = 72 km/h =72*5/18 m/s = 20 m/s

Time taken, t = 5 minutes = 300 s

Let a, s be acceleration and distance covered by the train.

Step 2: Formula used

v = u + at (First equation of motion)

$s=ut+\frac{1}{2}a{t}^2$ (Second equation of motion)

Where; u = Initial velocity, v= Final velocity, a = Acceleration, t= time, s= distance covered

Step 3: Finding the acceleration

v = u + at

$20=0+a\times 300$

$300a=20$

$a=\frac{20}{300}$

$a=\frac{1}{15}m{s}^{-2}$

Step 4: Finding the distance travelled

$s=ut+\frac{1}{2}a{t}^2$

$s=\left(0\right)t+\frac{1}{2}\left(\frac{1}{15}\right){\left(300\right)}^2$

$s=3000m$

$s=3km$

Hence, the distance travelled by train is 3 km.

Thus,

Option (C) 1/15 m/s2, 3 km is the correct option.

s=0×300+12×115×(300)2s=0\times 300+\frac{1}{2}\times\frac{1}{15}\times (300)^{2}