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 given values: $g = \frac{2 \times 2.50}{(0.74)^2}$ Calculating this yields: $g = \frac{5.00}{0.5476} \approx 9.12 \text{ m/s}²$

To find the average time t:

1. Calculate the total time: $0.74 + 0.69 + 0.81 = 2.24 ext{ s}$
2. Divide by the number of measurements: $\frac{2.24}{3} \approx 0.747 \text{ s}$ Thus, the average value of time t is 0.75 s (to two significant figures).

One way to improve their procedure is by using an electronic timer or a light gate/data logger to eliminate human reaction time, which can introduce errors in timing.

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

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