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 minutes into seconds, multiply the number of minutes by 60. Thus:

$6 ext{ minutes} imes 60 ext{ seconds/minute} = 360 ext{ seconds}$

Acceleration is calculated using the formula:

$a = \frac{v - u}{t}$ Where:

• $v = 25 ext{ m/s}$ (final velocity)
• $u = 0 ext{ m/s}$ (initial velocity, as the train starts from rest)
• $t = 360 ext{ s}$ (time in seconds)

Thus, the acceleration is:

$a = \frac{25 ext{ m/s} - 0 ext{ m/s}}{360 ext{ s}} = 0.069 ext{ m/s}^2$

The force can be calculated using Newton's second law of motion:

$F = ma$ Where:

• $m = 420000 ext{ kg}$ (mass of the train)
• $a = 0.069 ext{ m/s}^2$ (acceleration)

So the force required is:

$F = 420000 ext{ kg} \times 0.069 ext{ m/s}^2 = 29166.7 ext{ N}$

The distance travelled can be calculated using the formula:

$s = ut + \frac{1}{2}at^2$ Where:

• $u = 0 ext{ m/s}$ (initial velocity)
• $a = 0.069 ext{ m/s}^2$ (acceleration)
• $t = 360 ext{ s}$ (time)

Thus:

$s = 0 + \frac{1}{2} \times 0.069 ext{ m/s}^2 \times (360 ext{ s})^2 = 4500 ext{ m}$

During the 15-minute interval, the train maintains a constant speed of 25 m/s. The distance travelled can be computed as:

$d = vt$ Where:

• $v = 25 ext{ m/s}$ (constant speed)
• $t = 15 imes 60 ext{ s} = 900 ext{ s}$ (time in seconds)

Thus:

$d = 25 ext{ m/s} \times 900 ext{ s} = 22500 ext{ m}$

In the diagram, indicate the following forces:

• The thrust force ($F$) acting forward.
• The friction force ($F_r$) opposing the motion.
• The weight ($W$) of the train acting downwards.
• The normal reaction force acting upwards. Label these appropriately.

Constant speed means that the object covers equal distances in equal intervals of time without regard to direction. However, velocity includes both speed and direction. Therefore, if an object changes direction while maintaining the same speed, its velocity changes. For example, a car moving around a circular track travels at a constant speed but experiences a change in velocity due to its changing direction.

The speed-time graph will show:

• A straight line increasing from 0 m/s to 25 m/s over the first 6 minutes (360 seconds) during acceleration.
• A horizontal line at 25 m/s maintained from 6 minutes to 21 minutes, showing constant speed. The axes should be appropriately labelled: the vertical axis for speed and the horizontal axis for time.