A car travels along a straight level road - Leaving Cert Applied Maths - Question 1 - 2012

To find the deceleration, we use the formula:

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

Where:

• $v = 0$ (final speed when it stops)
• $u = 32 \, \text{m s}^{-1}$ (initial speed)
• $s = 128 \, \text{m}$ (distance)

Setting up the equation gives:

$0 = (32)^2 + 2a(128)$

Solving for $a$:

a = -\frac{32^2}{2(128)} = -4 \, \text{m s}^{-2}$$

To find the total distance traveled from P to Q, we need to analyze each segment:

1. Acceleration Phase:

   • Use the formula for distance: $s_1 = ut + \frac{1}{2}at^2$ Substituting: $s_1 = 8(12) + \frac{1}{2}(2)(12^2) = 240 \, \text{m}$

2. Constant Speed Phase:

   • Distance: $s_2 = vt = 32(7) = 224 \, \text{m}$

3. Deceleration Phase:

   • Distance is given as 128 m.

Finally, sum all the distances:

$|PQ| = s_1 + s_2 + s_3 = 240 + 224 + 128 = 592 \, \text{m}$

To find the speed when the car is 72 m from Q, we first need to identify the distance remaining after the car decelerates:

1. Distance from Q: Remaining distance is $72 \, m$. Total distance during deceleration is $128 \, m$.
2. Distance traveled during deceleration:
   • Distance traveled: $128 - 72 = 56 \, m$.

Using the formula: $v^2 = u^2 + 2as$ Where:

• $s = 56 \, m$
• $u = 32 \, m s^{-1}$
• $a = -4 \, m s^{-2}$

Setting up the equation: $v^2 = 32^2 + 2(-4)(56)$

Solving this yields: $v^2 = 1024 - 448 = 576$ Thus: $v = \sqrt{576} = 24 \, m s^{-1}$