FPM(2023) November Paper 02 Question 3/4 and solution

Question 3

A particle $P$ moves along the $x$-axis.

At time $t$ seconds ($t \geqslant 0$) the acceleration, $a$ m/s$^2$, of $P$ is given by $a = 6t - 16$. When $t = 0$, $P$ is at the origin and is moving with velocity $12$ m/s.

(a) Find an expression in terms of $t$ for

(i) the velocity of $P$ at time $t$ seconds

(ii) the displacement of $P$ at time $t$ seconds. (4)

(b) Hence find the time at which $P$ first returns to the origin. (3)

Solution

Step 1: Expression for velocity of $P$

The acceleration of the particle is given as:

$ a = 6t - 16. $

Velocity $v$ is obtained by integrating acceleration with respect to time:

$ v = \int a \, dt = \int (6t - 16) \, dt. $

Perform the integration:

$ v = 3t^2 - 16t + C, $

where $C$ is the constant of integration.

At $t = 0$, the velocity $v = 12 \, \mathrm{m/s}$. Substitute these values to find $C$:

$ 12 = 3(0)^2 - 16(0) + C \implies C = 12. $

Thus, the velocity of $P$ at time $t$ seconds is:

$ v = 3t^2 - 16t + 12. $

Step 2: Expression for displacement of $P$

The displacement $x$ is obtained by integrating velocity with respect to time:

$ x = \int v \, dt = \int (3t^2 - 16t + 12) \, dt. $

Perform the integration:

$ x = t^3 - 8t^2 + 12t + C_1, $

where $C_1$ is the constant of integration.

At $t = 0$, the particle is at the origin, so $x = 0$. Substitute these values to find $C_1$:

$ 0 = (0)^3 - 8(0)^2 + 12(0) + C_1 \implies C_1 = 0. $

Thus, the displacement of $P$ at time $t$ seconds is:

$ x = t^3 - 8t^2 + 12t. $

Step 3: Time at which $P$ first returns to the origin

The particle returns to the origin when $x = 0$. Solve the equation:

$ t^3 - 8t^2 + 12t = 0. $

Factorize the equation:

$ t(t^2 - 8t + 12) = 0. $

Solve for $t$:

$ t = 0 \quad \text{or} \quad t^2 - 8t + 12 = 0. $

Solve the quadratic equation $t^2 - 8t + 12 = 0$ using factorization:

$ t^2 - 8t + 12 = (t - 6)(t - 2) = 0. $

Thus:

$ t = 6 \quad \text{or} \quad t = 2. $

Since $t = 0$ corresponds to the initial position, the particle first returns to the origin at:

$ t = 2 \quad \text{seconds}. $

Final Answers

(a) (i) The velocity of $P$ at time $t$ seconds is:

$ v = 3t^2 - 16t + 12. $

(ii) The displacement of $P$ at time $t$ seconds is:

$ x = t^3 - 8t^2 + 12t. $

(b) The particle first returns to the origin at:

$ t = 2 \quad \text{seconds}. $

Question 4

(a) On the axes opposite, draw the line with equation

(i) $y = -x - 1$
(ii) $y - 3x + 8 = 0$
(iii) $2y = x + 8$

(b) Show, by shading on your graph, the region $R$ defined by the inequalities

$y \geqslant -x - 1$ and $y \geqslant 3x - 8$ and $2y \leqslant x + 8$. (1)

For all points in $R$ with coordinates $(x, y)$,

$$P = 2y - 3x.$$

(c) Find

(i) the greatest value of $P$

(ii) the least value of $P$. (4)

Solution

Step 1: Draw the lines

(a) On the axes, we draw the following lines:

(i) The line $y = -x - 1$

This line has a gradient of $-1$ and a $y$-intercept of $-1$. To plot this:

• Point 1: $(0, -1)$ (intercept)
• Point 2: $(1, -2)$ (substitute $x = 1$)

Join these points with a straight line.

(ii) The line $y - 3x + 8 = 0$

Rewrite as $y = 3x - 8$. This line has a gradient of $3$ and a $y$-intercept of $-8$. To plot this:

• Point 1: $(0, -8)$ (intercept)
• Point 2: $(1, -5)$ (substitute $x = 1$)

Join these points with a straight line.

(iii) The line $2y = x + 8$

Rewrite as $y = \frac{x}{2} + 4$. This line has a gradient of $\frac{1}{2}$ and a $y$-intercept of $4$. To plot this:

• Point 1: $(0, 4)$ (intercept)
• Point 2: $(2, 5)$ (substitute $x = 2$)

Join these points with a straight line.

Step 2: Identify the region $R$

The region $R$ is defined by the following inequalities:

• $y \geqslant -x - 1$ (above the blue line)
• $y \geqslant 3x - 8$ (above the red line)
• $2y \leqslant x + 8 \implies y \leqslant \frac{x}{2} + 4$ (below the green line)

Shade the intersection of these regions to represent $R$.

Step 3: Evaluate $P$ at vertices of $R$

The vertices of $R$ are determined by the intersections of the lines:

• Intersection of $y = -x - 1$ and $y = 3x - 8$
• Intersection of $y = -x - 1$ and $y = \frac{x}{2} + 4$
• Intersection of $y = 3x - 8$ and $y = \frac{x}{2} + 4$

Solve these equations to find the coordinates of the vertices.

Intersection 1: $y = -x - 1$ and $y = 3x - 8$

Equating the two equations:

$ -x - 1 = 3x - 8 $

$ -4x = -7 $

$ x = \frac{7}{4} $

$ y = -\frac{7}{4} - 1 = -\frac{11}{4} $

Vertex: $ \left(\frac{7}{4}, -\frac{11}{4}\right) $

Intersection 2: $y = -x - 1$ and $y = \frac{x}{2} + 4$

Equating the two equations:

$ -x - 1 = \frac{x}{2} + 4 $

$ -\frac{3x}{2} = 5 $

$ x = -\frac{10}{3} $

$ y = -\left(-\frac{10}{3}\right) - 1 = \frac{10}{3} - 1 = \frac{7}{3} $

Vertex: $ \left(-\frac{10}{3}, \frac{7}{3}\right) $

Intersection 3: $y = 3x - 8$ and $y = \frac{x}{2} + 4$

Equating the two equations:

$ 3x - 8 = \frac{x}{2} + 4 $

$ \frac{5x}{2} = 12 $

$ x = \frac{24}{5} $

$ y = \frac{24}{10} + 4 = \frac{12}{5} + 4 = \frac{32}{5} $

Vertex: $ \left(\frac{24}{5}, \frac{32}{5}\right) $

Step 4: Calculate $P = 2y - 3x$

Evaluate $P$ at each vertex:

• At $ \left(\frac{7}{4}, -\frac{11}{4}\right) $:
  $ P = 2\left(-\frac{11}{4}\right) - 3\left(\frac{7}{4}\right) = -\frac{22}{4} - \frac{21}{4} = -\frac{43}{4} = -10.75 $
• At $ \left(-\frac{10}{3}, \frac{7}{3}\right) $:
  $ P = 2\left(\frac{7}{3}\right) - 3\left(-\frac{10}{3}\right) = \frac{14}{3} + 10 = \frac{44}{3} $
• At $ \left(\frac{24}{5}, \frac{32}{5}\right) $:
  $ P = 2\left(\frac{32}{5}\right) - 3\left(\frac{24}{5}\right) = \frac{64}{5} - \frac{72}{5} = -\frac{8}{5} = -1.6 $

Therefore:

The greatest value of $P$ is $ \boxed{\frac{44}{3}}. $

The least value of $P$ is $ \boxed{-10.75}. $