1. (a) A particle falls from rest from a point P - Leaving Cert Applied Maths - Question 1 - 2012

To find the time t when the car and bus meet, we calculate their respective distances traveled:

1. Distance Traveled by the Car: The car starts from rest and accelerates:

   $v = u + at = 0 + 1t = t$

   The car reaches its maximum speed of 32 m s⁻¹:

   $t_{max} = \frac{32}{1} = 32\text{ seconds}$

   The distance after reaching maximum speed is:

   $s_{car} = \frac{1}{2}(t_{max})(32) = \frac{1}{2}(32)(32) = 512 m$

2. Distance Traveled by the Bus: The bus travels at 36 m/s for 12 seconds before deceleration starts:

   $s_{bus} = 36 \cdot 12 = 432 m$

   After that, it starts decelerating:

   Let time in seconds after the first 12 seconds be represented by t where the total time is $t + 12$:

   $s_{bus ext{ (after 12s)}} = 1914 - \frac{(36)(t)}{72} = distance\text{ covered by bus}$

3. Setting the Equations Equal: Setting the distances equal:

   $512 = \left(\frac{32}{2}(t + 12) + 432\right) o 0 = 616 - 19840 \to t = 40 \text{ seconds}$

Thus, we find:

t = 40 seconds.

1. Distance Traveled by the Car in 48 Seconds: The car accelerates during the first 32 seconds, reaching maximum speed:

   $s = ut + \frac{1}{2}at^2 = 32\cdot 32 = 256 m$

   After that, it travels at maximum speed for the remaining time:

   $total_{car} = 256 + 32(48-32) = 256 + 512 = 768 m$

2. Distance Traveled by the Bus in 48 Seconds: The bus travels at 36 m/s for 12 seconds, covering:

   $s_{bus} = 36 imes 12 = 432 m$

   After 12 seconds, it decelerates:

   Using:

   $s_{bus} = ut + \frac{1}{2} a t^2$

   The deceleration here is $0.75 m/s²$:

   The distance it covers from 12 seconds onward for 36 seconds is:

   $s = 36(48-12) + \frac{1}{2}(-0.75)(36)^2 = 936 - 486 = 450$

3. Total Distances After 48 Seconds: Finally, the total distance between them after 48 seconds is:

   $Distance = 768 - 450 = 318 m$

   Therefore, the distance between the car and the bus after 48 seconds is 318 m.