Past Exam Questions

(i) The acceleration is calculated using the formula:

\(a = \frac{v - u}{t}\)

where \(v = 16 \text{ m/s}\), \(u = 0 \text{ m/s}\), and \(t = 40 \text{ s}\).

Thus, \(a = \frac{16 - 0}{40} = 0.4 \text{ m/s}^2\).

(ii) The total distance is the area under the velocity-time graph, which is a trapezium. The area \(A\) is given by:

\(A = \frac{1}{2} \times (600 + T) \times 16 = 9040\)

Solving for \(T\):

\(9040 = \frac{1}{2} \times (600 + T) \times 16\)

\(9040 = 8 \times (600 + T)\)

\(1130 = 600 + T\)

\(T = 530 \text{ s}\)

(iii) The distance travelled while decelerating is the area of the triangle at the end of the graph:

\(s = \frac{1}{2} \times (600 - 530 - 40) \times 16\)

\(s = \frac{1}{2} \times 30 \times 16\)

\(s = 240 \text{ m}\)

(i) The velocity-time graph consists of three straight line segments: a positive gradient for 0 < t < 80, a zero gradient for 80 < t < 560, and a negative gradient for 560 < t < 600. The graph starts and ends on the t-axis at t = 0 and t = 600, respectively.

(ii) Using the area under the velocity-time graph for 0 < t < 80, we have:

\(\frac{1}{2} \times 80 \times v = 400\)

Solving for \(v\), we get \(v = 10 \text{ ms}^{-1}\).

(iii) The total distance is the area under the velocity-time graph. The area of the trapezium is:

\(D = \frac{1}{2} (600 + 480) \times 10 = 5400 \text{ m}\)

(iv) The acceleration for 0 < t < 80 is given by the gradient of the velocity-time graph:

\(v = 0 + at\)

\(10 = a \times 80\)

\(a = 0.125 \text{ ms}^{-2}\)

(a) For particle A, it moves with constant speed 8 m/s. Since B passes O 4 seconds after A, the displacement of A after t seconds is:

\(s_A = 8(4 + t) = 32 + 8t\)

For particle B, using the equation \(s = ut + \frac{1}{2}at^2\) with \(u = 20\) m/s and \(a = -2\) m/s², the displacement is:

\(s_B = 20t - \frac{1}{2} \times 2t^2 = 20t - t^2\)

(b) Set \(s_A = s_B\):

\(32 + 8t = 20t - t^2\)

Rearrange to form a quadratic equation:

\(t^2 - 12t + 32 = 0\)

Solving this gives \(t = 4\) and \(t = 8\).

(c) The graph for A is a straight line starting at \(s = 32\) when \(t = 0\) with a positive gradient. The graph for B is an inverted quadratic starting at the origin, reaching a maximum, and then returning to the axis. They intersect at \(t = 4\) and \(t = 8\).

(i) The velocity-time graph is a trapezium with three straight lines: a positive gradient for the first 3 seconds, a zero gradient for the next 6 seconds, and a negative gradient for the last 4 seconds. The time points 0, 3, 9, and 13 are shown on the time axis. The maximum velocity reached is 2.7 m/s.

(ii) To find the total distance, we calculate the area under the velocity-time graph, which is a trapezium. The formula for the area of a trapezium is:

\(\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}\)

Here, the bases are the time intervals: 6 s and 13 s, and the height is the maximum velocity 2.7 m/s. Thus, the total distance is:

\(\text{Total distance} = \frac{1}{2} \times (6 + 13) \times 2.7 = 25.65 \text{ m}\)

Alternatively, using the constant acceleration equations for each stage:

1. Stage 1: \(s_1 = 0.5 \times 0.9 \times 3^2 = 4.05 \text{ m}\)
2. Stage 2: \(s_2 = 2.7 \times 6 = 16.2 \text{ m}\)
3. Stage 3: \(s_3 = 0.5 \times (2.7 + 0) \times 4 = 5.4 \text{ m}\)

Total distance = \(s_1 + s_2 + s_3 = 4.05 + 16.2 + 5.4 = 25.65 \text{ m}\)

(i) The velocity-time graph consists of three segments: an upward sloping line from 0 to V over T₁ seconds, a horizontal line at V for T₂ seconds, and a downward sloping line from V to 0 over T₃ seconds. Using the equations of motion:

For acceleration: \(v = u + at \Rightarrow V = 0 + 0.3T_1 \Rightarrow T_1 = \frac{V}{0.3}\).

For deceleration: \(v = u - at \Rightarrow 0 = V - 1T_3 \Rightarrow T_3 = V\).

(ii) The total distance travelled is the area under the velocity-time graph, which is a trapezium:

\(S = \frac{1}{2} T_1 V + T_2 V + \frac{1}{2} T_3 V\).

Substitute \(T_1 = \frac{V}{0.3}\) and \(T_3 = V\):

\(S = \frac{1}{2} \left(\frac{V}{0.3}\right) V + T_2 V + \frac{1}{2} V^2\).

Given \(S = 12000\) m and \(T_1 + T_2 + T_3 = 552\):

\(12000 = \frac{1}{2} \left(\frac{V}{0.3}\right) V + T_2 V + \frac{1}{2} V^2\).

\(12000 = \frac{V^2}{0.6} + T_2 V + \frac{V^2}{2}\).

\(12000 = \frac{5V^2}{6} + T_2 V\).

\(T_2 = 552 - \frac{V}{0.3} - V\).

Substitute \(T_2\) into the equation:

\(12000 = \frac{5V^2}{6} + (552 - \frac{V}{0.3} - V)V\).

\(12000 = \frac{5V^2}{6} + 552V - \frac{V^2}{0.3} - V^2\).

\(12000 = \frac{5V^2}{6} + 552V - \frac{10V^2}{3} - V^2\).

\(12000 = 552V - \frac{13V^2}{6}\).

Multiply through by 6 to clear the fraction:

\(72000 = 3312V - 13V^2\).

Rearrange to form a quadratic equation:

\(13V^2 - 3312V + 72000 = 0\).

Solving this quadratic equation gives \(V = 24\).

(i) The velocity-time graph consists of three straight line segments:

• From \(t = 0\) to \(t = 8\), velocity increases from 0 to 8 m/s.
• From \(t = 8\) to \(t = 20\), velocity decreases from 8 m/s to 2 m/s.
• From \(t = 20\) to \(t = 26\), velocity increases from 2 m/s to 6.5 m/s.

(ii) The shaded region for \(20 \leq t \leq 26\) is the area under the velocity-time graph between these times.

(iii) To show \(s = 0.375t^2 - 13t + 202\):

1. Calculate \(s(20)\):

\(s(20) = \frac{1}{2} \times 8 \times 8 + \frac{1}{2} \times (8 + 2) \times 12 = 92 \text{ meters}.\)

2. Calculate acceleration in the third phase:

\(a = \frac{6.5 - 2}{6} = 0.75 \text{ m/s}^2.\)

3. Use the equation for displacement:

\(s(t) = 92 + 2(t - 20) + 0.375(t - 20)^2.\)

4. Simplify to get:

\(s(t) = 0.375t^2 - 13t + 202.\)

(i) The train accelerates from rest with an acceleration of 0.025 m s^-2 for 600 s. The velocity at the end of this period is given by:

\(v(600) = 0.025 \times 600 = 15 \text{ m s}^{-1}\)

The train then travels at a constant speed of 15 m s^-1 for 2600 s. Finally, it decelerates to rest with a deceleration of 0.0375 m s^-2. The time taken to decelerate to rest is given by:

\(0 = 15 - 0.0375 t_3\)

\(t_3 = \frac{15}{0.0375} = 400 \text{ s}\)

The total time is:

\(600 + 2600 + 400 = 3600 \text{ s}\)

(ii) The velocity-time graph consists of three segments: an upward slope for 600 s, a horizontal line for 2600 s, and a downward slope for 400 s. The distance is the area under the graph:

\(d = \frac{1}{2} \times 600 \times 15 + 2600 \times 15 + \frac{1}{2} \times 400 \times 15\)

\(d = 4500 + 39000 + 3000 = 46500 \text{ m}\)

(iii) The speed of the train is 7.5 m s^-1 at two points: during acceleration and deceleration. During acceleration:

\(7.5 = 0.025 t\)

\(t = \frac{7.5}{0.025} = 300 \text{ s}\)

During deceleration:

\(7.5 = 15 - 0.0375 (t - 3200)\)

\(0.0375 (t - 3200) = 7.5\)

\(t - 3200 = \frac{7.5}{0.0375} = 200\)

\(t = 3400 \text{ s}\)

(i) For 0 < t < 4, the acceleration is 0.75 m s^-2. The initial velocity is 0, so the velocity at t = 4 is given by:

\(v(4) = 0 + 0.75 \times 4 = 3 \text{ m s}^{-1}\)

For 4 < t < 54, the acceleration is 0, so the velocity remains constant at 3 m s^-1.

\(For 54 < t < 60, the acceleration is -0.5 m s^-2. The velocity at t = 60 is given by:\)

\(v(60) = v(54) + (-0.5) \times (60 - 54) = 3 + (-0.5) \times 6 = 0 \text{ m s}^{-1}\)

The velocity-time graph consists of three segments: a line with positive slope from (0,0) to (4,3), a horizontal line from (4,3) to (54,3), and a line with negative slope from (54,3) to (60,0).

(ii) The distance XY can be found by calculating the area under the velocity-time graph:

Area of the first segment (triangle):

\(\frac{1}{2} \times 4 \times 3 = 6 \text{ m}\)

Area of the second segment (rectangle):

\(50 \times 3 = 150 \text{ m}\)

Area of the third segment (triangle):

\(\frac{1}{2} \times 6 \times 3 = 9 \text{ m}\)

Total distance XY is:

\(6 + 150 - 9 = 165 \text{ m}\)

(i) Using the equation \(s = \frac{1}{2} (u + v) t\), where \(u = 1.4 \, \text{m/s}\), \(v = 1.1 \, \text{m/s}\), and \(t = 1.2 \, \text{s}\):

\(s = \frac{1}{2} (1.4 + 1.1) \times 1.2 = 1.5 \, \text{m}\)

To find \(d\), use \(v = u + at\):

\(1.1 = 1.4 + (-d) \times 1.2\)

\(d = 0.25 \, \text{m/s}^2\)

(ii) For the motion from \(B\) to \(C\), use \(0 = u^2 + 2(-d)s\):

\(0 = u^2 + 2(-0.25) \times 2\)

\(u = 1 \, \text{m/s}\)

For time \(t\), use \(s = 0 - \frac{1}{2}(-d)t^2\):

\(2 = 0 - \frac{1}{2}(-0.25)t^2\)

\(t = 4 \, \text{s}\)

(iii) The velocity-time graph consists of a line joining \((0, 1.4)\) and \((1.2, 1.1)\), and another line joining \((1.2, -1)\) and \((5.2, 0)\).

(i) The velocity-time graph consists of three straight line segments:

• The first segment has a positive slope, starting from 0 and reaching V at time t₁.
• The second segment is a horizontal line at velocity V from t₁ to t₁ + 600 s.
• The third segment has a negative slope, decreasing from V to 0 at time 1,000 s.

(ii) Using the area under the graph to represent the distance:

For the entire journey: \(\frac{1}{2} (600 + 1000) V = 20000\)

Solving for \(V\):

\(V = 25 \text{ m/s}\)

(iii) Given acceleration \(a = 0.15 \text{ m/s}^2\), find \(t_1\):

\(V = a t_1 \Rightarrow t_1 = \frac{V}{0.15} = \frac{25}{0.15} = 166.6 \text{ s}\)

Time for third stage \(t_3 = 1000 - (t_1 + 600) = 233.3 \text{ s}\)

Distance during third stage \(s_3 = \frac{1}{2} \times 233.3 \times 25 = 2920 \text{ m}\)

(a) To find the time for which the car is accelerating, use the formula for distance under constant acceleration:

\(s = \frac{(u+v)}{2} t\)

where \(s = 200\) m, \(u = 0\) m/s, \(v = 25\) m/s. Solving for \(t\):

\(200 = \frac{(0+25)}{2} t\)

\(200 = 12.5t\)

\(t = 16\) s

(b) The velocity-time graph is a trapezium with vertices at (0, 0), (16, 25), (32, 25), and (37, 0).

(c) The total distance traveled is the sum of the distances during acceleration, constant speed, and deceleration:

Total distance = \(200 + 400 + \frac{1}{2} \times 25 \times 5 = 662.5\) m

The total time is \(16 + 16 + 5 = 37\) s.

The average speed is:

\(\frac{662.5}{37} = 17.9\) m/s (3 s.f.)

(a) The velocity-time graph is a trapezium. The car accelerates from rest to a constant velocity, travels at this constant velocity, and then decelerates to rest. The gradient of the right-hand side is approximately twice that of the left-hand side due to the deceleration being \(2a\).

(b) To find the constant velocity, use the formula: \(v = \frac{500}{25} = 20 \text{ m s}^{-1}\).

Using the equation \(v^2 = u^2 + 2as\), where \(u = 0\), \(v = 20\), and \(s = 50\):

\(20^2 = 0 + 2a \times 50\)

\(400 = 100a\)

\(a = 4\)

(c) Time to accelerate: \(t = \frac{v}{a} = \frac{20}{4} = 5 \text{ s}\).

Deceleration time: \(t = \frac{v}{2a} = \frac{20}{8} = 2.5 \text{ s}\).

Total time = \(5 + 25 + 2.5 = 32.5 \text{ s}\).

(a) The velocity-time graph is a trapezium with the deceleration steeper than the acceleration. The total time from start to stop is 200 s.

(b) The total distance travelled is given by the area under the velocity-time graph, which is a trapezium. The area is \(0.5 \times (170 + 200) \times V = 2775\). Solving for \(V\), we get:

\(0.5 \times 370 \times V = 2775\)

\(185V = 2775\)

\(V = 15 \text{ m s}^{-1}\)

(c) The acceleration \(a\) can be found using the formula \(a = \frac{V}{t}\), where \(t = 20 \text{ s}\). Thus:

\(a = \frac{15}{20} = 0.75 \text{ m s}^{-2}\)

The motion of the bus can be divided into three phases:

1. Acceleration from rest for 5 s at 2.1 \(\text{m/s}^2\):

   The final velocity \(v = u + at = 0 + 2.1 \times 5 = 10.5 \text{ m/s}\).

2. Constant speed for 24 s:

   The velocity remains 10.5 \(\text{m/s}\).

3. Deceleration to stop in 6 s:

   The initial velocity for this phase is 10.5 \(\text{m/s}\) and the final velocity is 0 \(\text{m/s}\).

The velocity-time graph is a trapezium with vertices at \((0, 0)\), \((5, 10.5)\), \((29, 10.5)\), and \((35, 0)\).

The area under the velocity-time graph gives the total distance traveled:

Area = \(\frac{1}{2} \times (24 + 35) \times 10.5\) or \(\frac{1}{2} \times 5 \times 10.5 + 24 \times 10.5 + \frac{1}{2} \times 6 \times 10.5\).

Calculating the area:

Area = \(\frac{1}{2} \times 59 \times 10.5 = 309.75 \text{ m}\).

Thus, the distance between the two bus stops is 309.75 m or 310 m.

(i) The runner accelerates from rest at 1.2 m/s² for 5 s, reaching a velocity of:

\(v = 1.2 \times 5 = 6 \text{ m/s}\)

The runner then moves at this constant speed for 12 s, and finally decelerates uniformly to rest over 3 s.

The velocity-time graph consists of three segments: a linearly increasing segment, a constant segment, and a linearly decreasing segment.

The total distance is the area under the velocity-time graph, which is a trapezium:

\(\text{Area} = \frac{1}{2} \times (12 + 20) \times 6 = 96 \text{ m}\)

(ii) The cyclist accelerates uniformly for 10 s and decelerates uniformly to rest at Q at t = 30. The total time for the cyclist is 20 s, and the distance is 96 m.

Using the formula for the area of a triangle:

\(\frac{1}{2} \times 20 \times v = 96\)

\(v = 9.6 \text{ m/s}\)

Using the equation \(s = ut + \frac{1}{2} a t^2\) with \(u = 0\), \(s = 48\), and \(t = 10\):

\(48 = \frac{1}{2} a (10)^2\)

\(a = \frac{9.6}{10} = 0.96 \text{ m/s}^2\)

(i) To find the distance travelled in the first 16 s, we divide the motion into two parts: acceleration and constant speed.

For the first 6 s, the sprinter accelerates from rest to 12 m/s. The distance \(s_1\) is given by:

\(s_1 = \frac{1}{2} (0 + 12) \times 6 = 36 \text{ m}\)

For the next 10 s, the sprinter travels at a constant speed of 12 m/s. The distance \(s_2\) is:

\(s_2 = 10 \times 12 = 120 \text{ m}\)

Thus, the total distance for the first 16 s is:

\(36 + 120 = 156 \text{ m}\)

The displacement-time graph is a curve concave up for \(0 < t < 6\) starting at (0, 0) and ending at (6, 36), a straight line with positive gradient from (6, 36) to (16, 156), and a curve concave down from (16, 156) to (20, 200).

(ii) To find \(V\), use the equation for the final 4 s of deceleration:

\(44 = \frac{1}{2} (12 + V) \times 4\)

Solving for \(V\):

\(44 = 2(12 + V)\)

\(44 = 24 + 2V\)

\(20 = 2V\)

\(V = 10\)

(i) The velocity-time graph is a trapezium. The train accelerates to 25 m/s, travels at constant speed, then decelerates to rest. The right-hand slope is steeper than the left-hand slope because the deceleration is twice the acceleration.

(ii) Let the acceleration be a. Then, the deceleration is 2a. The time to decelerate from 25 m/s to 0 is T/2. The total time is 180 s, so the time at constant speed is:

\(180 - T - \frac{T}{2} = 180 - 1.5T\)

(iii) The total distance is 3300 m. Using the area under the velocity-time graph, the distance is:

\(0.5 \times [180 + (180 - 1.5T)] \times 25 = 3300\)

Solving for T:

\(0.5 \times (360 - 1.5T) \times 25 = 3300\)

\(360 - 1.5T = 264\)

\(1.5T = 96\)

\(T = 64\)

The time to decelerate is 32 s. The distance while decelerating is:

\(0.5 \times 32 \times 25 = 400 \text{ m}\)

(i) The car accelerates from rest with an acceleration of 3 m/s² for 10 s. The final velocity after 10 s is given by:

\(v = 3 \times 10 = 30 \text{ m/s}\)

The car then travels at this constant speed for 30 s. The distance travelled during this time is:

\(s = \frac{1}{2} (30 + 40) \times 30 = 1050 \text{ m}\)

Thus, the total distance travelled in the first 40 s is 1050 m.

(ii) The car decelerates uniformly to rest, covering a distance of 450 m in this stage. The time taken for this stage is:

\(\text{Time taken} = \frac{450}{15} = 30 \text{ s}\)

The total time of motion for the car is 70 s. The motorcycle takes 50 s to travel 1500 m. Using the equation:

\(1500 = \frac{1}{2} (30 + 50) \times V\)

or

\(1500 = 30V + 0.5 \times 20V\)

Solving gives \(V = 37.5 \text{ m/s}\).

The motorcycle accelerates for 5 s and decelerates for 15 s. The acceleration \(a\) is:

\(a = \frac{37.5}{5} = 7.5 \text{ m/s}^2\)

(iii) The displacement-time graph should show three stages: acceleration, constant speed, and deceleration, with correct curvature and labels at \(t = 10, 40, 70 \text{ s}\) corresponding to displacements of 150, 1050, and 1500 m respectively.

(i) The area representing the displacement of P from X at the instant when Q starts is a composite figure consisting of two rectangles. The first rectangle has boundaries at t = 0 and t = 20, with v = 0 and v = 2.5. The second rectangle has boundaries at t = 20 and t = T, with v = 0 and v = 4.

(ii) The displacement of P when Q starts is given by:

\(50 + 4(T - 20) = 70\)

Simplifying gives:

\(4T - 30 = 70\)

\(4T = 100\)

\(T = 25\)

(iii) The distance between P and Q when Q's speed reaches 4 m s^-1 is calculated by:

\(\text{Distance} = 70 + (4 - 2.5) \times 20\)

\(\text{Distance} = 70 + 30 = 100 \text{ m}\)

(iv) For the displacement-time graph, P is represented by two straight line segments: the first with a positive slope and the second with a steeper slope, with t = 20 indicated. For Q, the first and second segments are parallel to P's and displaced to the right, with t = 25 and t = 45 indicated.

(i) To find the distance AC, calculate the area under the velocity-time graph from t = 0 to t = 10 s. The velocity is 7 m/s for 10 seconds, so:

\(AC = 7 \times 10 = 70 \text{ m}\)

To find the distance AB, calculate the net displacement by subtracting the area from t = 15 to t = 30 s (where velocity is -4 m/s) from the distance AC:

\(BC = 4 \times 15 = 60 \text{ m}\)

\(AB = AC - BC = 70 - 60 = 10 \text{ m}\)

(ii) The graph of x against t consists of three connected straight line segments:

1. From t = 0 to t = 10, x increases linearly from 0 to 70 m.
2. From t = 10 to t = 15, x remains constant at 70 m.
3. From t = 15 to t = 30, x decreases linearly from 70 m to 10 m.

The graph should show:

• At t = 10, x = 70 m.
• At t = 15, x = 70 m.
• At t = 30, x = 10 m.