The graph shows the speed, v metres per second (m/s), of a car at time t seconds - OCR - GCSE Maths - Question 15 - 2017 - Paper 1

To check if the car accelerated from 0 to 60 miles per hour (mph), first convert 60 mph to metres per second:

60 ext{ mph} = 60 imes rac{1609.34 ext{ m}}{3600 ext{ s}} \approx 26.82 ext{ m/s}.

Now, observe the graph. By the 10-second mark, the speed is around 28 m/s. Thus, the graph supports the claim as the speed exceeds 26.82 m/s in the given time period.

The acceleration can be estimated by drawing a tangent line at t = 7 seconds and calculating the slope. The speed changes from approximately 20 m/s (at t = 6) to about 25 m/s (at t = 7), over 1 second. Therefore, the approximate acceleration is:

$a = \frac{v_f - v_i}{t_f - t_i} = \frac{25 ext{ m/s} - 20 ext{ m/s}}{1 ext{ s}} = 5 ext{ m/s}^2.$

Since the speed, v, is directly proportional to the square of the time, t, we express this relationship mathematically as:

$v = k t^2$

where k is a proportionality constant.

The graph indicates that after 10 seconds, the speed is leveling off, suggesting that the car will not continue to accelerate indefinitely. Therefore, Georgina's statement is incorrect.