(a) A car is travelling at a uniform speed of 14 ms^-1 when the driver notices a traffic light turning red 98 m ahead - Leaving Cert Applied Maths - Question 1 - 2010

In this scenario, the car continues to travel for 1 second at the initial speed. The distance covered during this time can be calculated as:

$s = ut = 14 imes 1 = 14 ext{ m}$

Now, the new distance to the traffic light is:

$s' = 98 - 14 = 84 ext{ m}$

Using the same equation of motion:

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

where u = 14 ms^-1 and v = 0:

$0^2 = (14)^2 + 2a(84)$

This simplifies to:

$0 = 196 + 168a$

Solving for a gives:

a = -rac{196}{168} = -rac{7}{6} ext{ ms}^{-2}

Thus, the minimum constant deceleration required is approximately 1.17 ms^-2.

Let x be the distance from P to Q.

In the first second, the particle moves 25 m:

$PX = 25 m$

In the last 3 seconds, it travels rac{13}{20}x:

Now applying the equation of motion:

For the entire distance from P to Q: |PQ| = ut + rac{1}{2}at^2

Here, let the distance from P to Q be represented as x:

$|PQ| = 20(4) + 5a(4)$

From the earlier information: x = 25 + rac{1}{20}(\frac{13}{20}x)

On solving and substituting the values: $x = 300 ext{ m}$ Therefore, the total distance from P to Q is 300 m.