A driver is road-testing two minibuses, A and B, for a taxi company - AQA - A-Level Maths Pure - Question 15 - 2018 - Paper 2

To find the time for minibus B, we follow similar steps:

1. The acceleration for B is given by: $a_B = 0.024 t^3$

2. Integrating the acceleration, we find the velocity: $v_B = \int 0.024 t^3 dt = 0.006 t^4 + c$

   At t = 0, v_B = 0, hence ( c = 0 ).

   Thus,: $v_B = 0.006 t^4$

3. Next, we integrate the velocity to find the displacement: $s_B = \int v_B dt = \int 0.006 t^4 dt = 0.0012 t^5 + k$

   Again, at t = 0, s_B = 0, so ( k = 0 ).

   Therefore,: $s_B = 0.0012 t^5$

4. To find the time taken for B to travel 100 metres, we set s_B = 100: $0.0012 t^5 = 100$

   Solving for t gives: $t = (\frac{100}{0.0012})^{1/5} = 9.642$ seconds.

Comparing the times:

• Time taken for minibus A: 9.657 seconds.
• Time taken for minibus B: 9.642 seconds.

Since 9.642 seconds (B) is less than 9.657 seconds (A), the company should choose minibus B.

The models assume that both minibuses start moving immediately when t = 0. However, in reality, there may be a reaction time for each driver before they start accelerating. If Driver B's reaction time is significantly greater than Driver A's, then in practice, Driver A could potentially travel 100 metres faster than Driver B despite what the model suggests. This discrepancy may lead the company to make a decision based on incomplete data regarding the drivers' actual performance.