A car travels from p to q along a straight level road - Leaving Cert Applied Maths - Question 1 - 2007

Using the formula for acceleration:

$a = \frac{v - u}{t}$

Where:

• v = final velocity = 15 m/s
• u = initial velocity = 0 m/s (starts from rest)
• t = time = 5 s

Substituting the values:

$a = \frac{15 - 0}{5} = 3 \text{ m/s}^2$

Thus, the uniform acceleration is 3 m/s².

Using the same approach for deceleration:

$a = \frac{v - u}{t}$

Where:

• v = final velocity = 0 m/s (comes to rest)
• u = initial velocity = 15 m/s
• t = time = 3 s

Substituting the values:

$a = \frac{0 - 15}{3} = -5 \text{ m/s}^2$

Thus, the uniform deceleration is 5 m/s².

To find the total distance traveled:

1. Distance during acceleration: $d_1 = \frac{1}{2} a t^2 = \frac{1}{2} \times 3 \times (5^2) = \frac{1}{2} \times 3 \times 25 = 37.5 \text{ m}$

2. Distance at constant speed: $d_2 = speed \times time = 15 \times 20 = 300 \text{ m}$

3. Distance during deceleration: $d_3 = \frac{1}{2} (u + v) t = \frac{1}{2} (15 + 0) \times 3 = \frac{1}{2} \times 15 \times 3 = 22.5 \text{ m}$

Total distance:

$|pq| = d_1 + d_2 + d_3 = 37.5 + 300 + 22.5 = 360 \text{ m}$

To find the speed of the car at 13.5 m:

Since 13.5 m is within the distance covered during the acceleration phase:

Using the equation:

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

Where:

• u = 0 m/s (initially at rest)
• a = 3 m/s² (acceleration)
• s = 13.5 m

Substituting the known values:

$v^2 = 0 + 2 \times 3 \times 13.5$

Calculating:

$v^2 = 81 \implies v = 9 \text{ m/s}$

Thus, the speed of the car when it is 13.5 metres from p is 9 m/s.