The symbol 'g' can be used to refer to the acceleration due to gravity - Edexcel - GCSE Physics Combined Science - Question 4 - 2018 - Paper 1

Using the formula: $g = \frac{2h}{t^2}$ Substituting the values:

• h = 2.50 m
• t = 0.74 s

Calculating: $g = \frac{2 \times 2.50}{(0.74)^2} = \frac{5.00}{0.5476} \approx 9.11 \text{ m/s}^2$

The average time t can be calculated from: $\text{average} = \frac{0.74 + 0.69 + 0.81}{3} = \frac{2.24}{3} \approx 0.747 \text{ s}$ Rounding to two significant figures, the average value of time t is 0.75 s.

The students could use an electronic timer or data logger to eliminate reaction time errors and improve measurement accuracy.

Using the equation: $a = \frac{(v^2 - u^2)}{2s}$ Where:

• v = 0 m/s (final speed)
• u = 15 m/s (initial speed)
• s = 14 m (distance)

Substituting the values: $a = \frac{(0^2 - 15^2)}{2 \times 14} = \frac{-225}{28} \approx -8.04 ext{ m/s}^2$ Thus, the deceleration of the car is approximately 8.04 m/s².