1. (a) A particle P, of mass 3 kg, is projected along a rough inclined plane from the point A with speed 4 m s⁻¹ - Leaving Cert Applied Maths - Question 1 - 2019

To find the distance |AB|, we use the kinematic equation:

$v^2 = u^2 + 2as$

Where:

• Initial velocity, ( u = 4 ) m/s,
• Final velocity, ( v = 0 ) m/s,
• Acceleration, ( a = -\frac{15g}{41} )

Plugging the values in:

$0 = 4^2 + 2\left(-\frac{15g}{41}\right)s$

Solving for ( s ) gives:

$|AB| = 2.46 \text{ m}$

As particle P slides back down, we again apply the kinematic equation. After sliding back down through point A, its final speed can be determined using:

$v^2 = u^2 + 2as$

where:

• Initial speed (at B), ( u = 0 )
• Acceleration (acting down the plane), ( a = \frac{15g}{41} )
• The distance calculated previously, ( s = 2.46 ) m

Hence,

$v^2 = 0 + 2 \cdot \frac{15g}{41} \cdot 2.46$

Calculating this gives:

$v \approx 1.88 \text{ m/s}$

To determine the distance between Maynooth and Leixlip, we calculate the distance traveled by train A:

Using the formula for distance under uniform acceleration:

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

Here:

• Initial velocity ( u = 0 ),
• Time ( t = 30 ) seconds at ( a = 0.5 , m/s^2 )

Therefore:

$s = 0 + \frac{1}{2} \cdot 0.5 \cdot (30)^2$

this results in:

$s = 720 m$

Thus, the distance between Maynooth and Leixlip is ( 720 \text{ m} ).

To find the meeting point, we must calculate when both trains have traveled equal distances. Train A accelerates then travels at a constant speed, while Train B starts at a constant speed and begins decelerating after 3 minutes. Let:

• ( t ) be the time train A travels,
• For Train A, after accelerating for ( t_a ):

$s_A = \frac{1}{2} \cdot a_A \, t_a^2 + v_A \, t_b$

For Train B:

• If it travels for ( t_b = t - 180 ):

$s_B = v_B \, t_b - \frac{1}{2} \, d \, t_b^2$

Setting ( s_A = s_B ) will allow solving for ( t ).

To find the deceleration of train A when it stops at Leixlip:

Using the known distance and time formula:

Using the distance covered by train A:

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

Substituting to find ( a ) when it decelerates:

• Where ( s = 720 m ), ( u = 25 , m/s ), and after traveling for ( t = 336 ) seconds.

Thus leads to: $a = \frac{(720) - (u)t}{\frac{1}{2}t^2}$

This provides the deceleration as ( 0.54 , m/s^2 ).