A train of mass 420000 kg started from rest and accelerated to a velocity of 25 m s^-1 in a time of 6 minutes - Leaving Cert Physics - Question 7 - 2022

To convert 6 minutes to seconds, multiply by 60:

6 ext{ min} imes 60 rac{ ext{sec}}{ ext{min}} = 360 ext{ sec}

Acceleration can be calculated using the formula:

a = rac{v - u}{t}

Where:

• $v = 25 ext{ m s}^{-1}$ (final velocity)
• $u = 0 ext{ m s}^{-1}$ (initial velocity)
• $t = 360 ext{ s}$ (time in seconds)

Substituting the values:

a = rac{25 - 0}{360} = 0.069 ext{ m s}^{-2}

Using Newton's second law, the force can be calculated as:

$F = m imes a$ Where:

• $m = 420000 ext{ kg}$
• $a = 0.069 ext{ m s}^{-2}$

So,

$F = 420000 imes 0.069 = 29166.7 ext{ N}$

The distance can be calculated using the formula:

s = ut + rac{1}{2} a t^2

Substituting the known values:

• $u = 0$
• $a = 0.069 ext{ m s}^{-2}$
• $t = 360 ext{ s}$

Then,

s = 0 + rac{1}{2} imes 0.069 imes (360)^2 = 4500 ext{ m}

During the 15 minutes at constant speed, the distance can be calculated as:

$s = vt$ Where:

• $v = 25 ext{ m s}^{-1}$
• $t = 15 ext{ min} = 15 imes 60 ext{ sec} = 900 ext{ sec}$$

Thus,

$s = 25 imes 900 = 22500 ext{ m}$

A labelled diagram would typically show the following forces:

• Weight (W) acting downwards.
• Normal force (N) acting upwards.
• Thrust (F) acting forward.
• Frictional force (Fr) acting backwards, balanced by thrust when moving at constant speed.

An object can maintain a constant speed while changing direction, which would result in a change in velocity. Velocity is a vector quantity, meaning it has both magnitude and direction. If the direction changes while the speed remains the same, the overall velocity changes.

The speed-time graph would consist of:

• A straight line increasing from 0 to 25 m/s over the first 6 minutes (360 seconds) representing acceleration.
• A horizontal line at 25 m/s for the next 15 minutes (900 seconds), showing constant speed.
• Then, a horizontal line at 0 speed after 21 minutes, as the journey ends.