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

First, convert 60 miles per hour to metres per second. Since 1 mile = 1.6 km,

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

Now, observe the graph for the first 10 seconds. The maximum speed at t = 10 is about 30 m/s, which is greater than 26.82 m/s. Thus, the graph supports the claim that the car has accelerated to a speed of at least 60 mph in the first 10 seconds.

To estimate the acceleration at t = 7 seconds, draw a tangent line at t = 7 on the graph. Measure the rise and run of the tangent line. If the tangent line rises by approximately 16 m/s over a run of 2 seconds, the acceleration a can be calculated by the formula:

$a = \frac{\Delta v}{\Delta t} = \frac{16}{2} = 8 \text{ m/s}^2.$

Given that the speed v is directly proportional to the square of the time t, we can express this as:

$v = k t^2,$

where k is a constant. To determine the value of k, we can substitute a known value from the graph. For example, at t = 2 seconds, if v = 4 m/s, we substitute:

$4 = k (2^2) \Rightarrow k = 1.$

Thus, $v = 1 t^2.$

The graph approaches a maximum speed and begins to level off at around t = 10 seconds. Therefore, it is incorrect to say that the speed of the car will continue to increase without bound after 10 seconds; instead, it approaches a terminal speed.