A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming the acceleration isuniform, nd(i) the acceleration.(ii) the distance traveled by the train for attaining this velocity.

(i) Acceleration is defined as the rate of change of velocity per unit of time. In this case, the train changes its velocity from 0 km/h to 72 km/h in 5 minutes.

First, we need to convert the units to be consistent. Velocity should be in m/s and time in seconds.

72 km/h = 72 * 1000 m/3600 s = 20 m/s

5 minutes = 5 * 60 = 300 seconds

Now, we can use the formula for acceleration:

a = Δv / Δt

a = (v_f - v_i) / (t_f - t_i)

where: v_f = final velocity = 20 m/s v_i = initial velocity = 0 m/s (since the train starts from rest) t_f = final time = 300 s t_i = initial time = 0 s (at the start)

So,

a = (20 m/s - 0 m/s) / (300 s - 0 s) = 20 m/s / 300 s = 0.067 m/s²

(ii) The distance traveled by the train can be found using the equation of motion:

d = v_it + 0.5a*t²

where: d = distance v_i = initial velocity = 0 m/s (since the train starts from rest) a = acceleration = 0.067 m/s² t = time = 300 s

So,

d = 0 m/s * 300 s + 0.5 * 0.067 m/s² * (300 s)² = 0 + 0.5 * 0.067 m/s² * 90000 s² = 3015 m

So, the train travels a distance of 3015 meters to attain this velocity.