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 to seconds, multiply by 60. Therefore, 6 minutes = 6 × 60 = 360 seconds.

Acceleration can be calculated using the formula: $a = \frac{\Delta v}{\Delta t}$ where $\Delta v = v - u$ Given that the initial velocity (u) is 0 and the final velocity (v) is 25 m/s:

during a time duration of 360 seconds:

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

Using Newton's second law: $F = m \cdot a$ where: m = 420000 kg and a = 0.069 m/s²:

$F = 420000 \cdot 0.069 = 29166.7 \, \text{N}$

Distance can be calculated using the formula: $s = ut + \frac{1}{2}at^2$. As the initial velocity (u) is 0, we have:

$s = 0 + \frac{1}{2} \cdot 0.069 \cdot (360)^2 = 4500 \, \text{m}$

The train maintains a constant speed of 25 m/s for 15 minutes. To find the distance: Distance = speed × time. 15 minutes = 15 × 60 = 900 seconds, therefore:

Distance = 25 m/s × 900 s = 22500 m.

A labelled diagram should show:

• The gravitational force (weight) acting downward (W)
• The normal force acting upward (N)
• The forward driving force (F) acting horizontally
• The frictional force (F_r) acting in the opposite direction of motion.

An object can have a constant speed if it travels the same distance in equal intervals of time. However, velocity takes into account direction. If the object changes direction while maintaining the same speed, its velocity changes, meaning it is not constant.

The speed-time graph will have:

• The x-axis labelled as "time (s)"
• The y-axis labelled as "speed (m/s)"

From 0 to 360 seconds, the graph will show a linear increase from 0 to 25 m/s. From 360 seconds to 1260 seconds (21 minutes), the graph will remain constant at 25 m/s.