Two cars, A and B, start from rest at O and begin to travel in the same direction - Leaving Cert Applied Maths - Question 10 - 2015

To find the distance traveled by each car after 4 seconds, we need to integrate their speed functions from 0 to 4 seconds:

1. For car A:

   $S_A = \int_{0}^{4} t^2 \, dt = \left[ \frac{t^3}{3} \right]_{0}^{4} = \frac{4^3}{3} = \frac{64}{3} \, \text{m}$

2. For car B:

   $S_B = \int_{0}^{4} (6t - 0.5t^2) \, dt = \left[ 3t^2 - \frac{t^3}{6} \right]_{0}^{4} = 3(4^2) - \frac{(4^3)}{6} = 48 - \frac{64}{6} = \frac{112}{3} \, \text{m}$

The distance between the cars is given by:

$S_B - S_A = \frac{112}{3} - \frac{64}{3} = \frac{48}{3} = 16 \, \text{m}$

To create the sketch, draw a graph with time on the x-axis (0 to 4 seconds) and speed on the y-axis.

• Plot the speed of car A as a curve representing $v_A = t^2$.
• Plot the speed of car B as a curve representing $v_B = 6t - 0.5t^2$.
• Shade the area between the two curves up to 4 seconds, identifying this area as the distance between the cars after 4 seconds.
  This representation visually demonstrates the speeds of both cars and the distance between them.

To find the cost function, we integrate the marginal cost function $M(x)$:

1. Starting from the marginal cost $M(x) = 74 + 1.1x + 0.03x^2$, we integrate:

   $C(x) = \int M(x) \, dx = \int (74 + 1.1x + 0.03x^2) \, dx$

   This gives us:

   $C(x) = 74x + 0.55x^2 + 0.01x^3 + F$

   where F is the constant of integration.

To find the increase in cost, we evaluate $C(160)$ and $C(120)$:

1. Calculate $C(160)$:

   $C(160) = 74(160) + 0.55(160)^2 + 0.01(160)^3 + F$

   $C(160) = 11840 + 1408 + 256 + F = 13404 + F$

2. Calculate $C(120)$:

   $C(120) = 74(120) + 0.55(120)^2 + 0.01(120)^3 + F$

   $C(120) = 8880 + 790 + 144 + F = 9814 + F$

3. The increase in cost is given by:

   $C(160) - C(120) = (13404 + F) - (9814 + F) = 13404 - 9814 = 3588$

Given that $C(10) = 3500$, we substitute back into the cost function:

1. First, calculate $C(10)$:

   $C(10) = 74(10) + 0.55(10)^2 + 0.01(10)^3 + F$

   $C(10) = 740 + 55 + 1 + F = 796 + F$

2. Setting this equal to 3500:

   $796 + F = 3500$

3. Solving for F:

   $F = 3500 - 796 = 2704$

Thus, the fixed costs are F = 2704.