8.1 The distance (D), in metres, covered by a car in t seconds, before reaching a certain point on a straight road, is given by the equation D(t) = -0.5t^2 + 20t Determine: 8.1.1 The distance covered by the car in t = 10 seconds - NSC Technical Mathematics - Question 8 - 2020 - Paper 1

To find the velocity of the car at t = 12 seconds, we need to first find the derivative of the distance function:

$D'(t) = -t + 20$

Now, substitute t = 12 into the derivative:

$D'(12) = -(12) + 20 = 8$

Thus, the velocity of the car at t = 12 seconds is 8 m/s.

The total surface area (TSA) is given by:

$TSA = 2(\text{area of base}) + (\text{perimeter of base}) × y$

For the triangular base with lengths 3 cm, 4 cm, and 5 cm, the area is:

$\text{area} = \frac{1}{2} \times 3 \times 4 = 6 cm^2$

The perimeter is:

$\text{perimeter} = 3 + 4 + 5 = 12 cm$

Inserting into the TSA formula:

$3600 = 2(6) + 12y$ $3600 = 12 + 12y$ $12y = 3600 - 12$ $12y = 3588$ $y = \frac{3588}{12} = 299$

However, to express y in terms of x, derive the previous result to show: $y = \frac{300 - x^2}{x}$

The volume of the prism can be computed using:

$V = \text{Area of base} \times y$

Substituting the area of the triangular base:

$V = 6y$

From the previous result, replace y to get: $V = 6 \left( \frac{300 - x^2}{x} \right)$

This simplifies to: $V = \frac{1800 - 6x^2}{x}$

To maximize the volume, take the derivative of V with respect to x:

$\frac{dV}{dx} = \frac{(300 - x^2)(-6) - (1800 - 6x^2)(-1)}{x^2}$

Set this derivative to zero:

$-1800 + 6x^2 = 0$

Solving this yields:

$x^2 = 300$ $x = \sqrt{300} = 10$

Thus, the value of x that will maximise the volume of the block is 10 cm.