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TuitionGoWhere Exam Practice (AI) - Physics H1 A-Level

PRACTICE PAPER - VERSION 5

Subject: Physics
Level: H1 (8867)
Paper: Practice Paper (Mechanics Focus)
Duration: 1 hour 15 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your name, class, and date in the spaces provided.
Answer all questions.
You are advised to spend approximately 15 minutes on Section A and 60 minutes on Section B.
All working must be clearly shown.
Use $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.

Section A: Structured Questions

Answer all questions in this section. Each question carries marks as indicated.

1. State the principle of conservation of linear momentum.
[2]

2. A car of mass $1200 \text{ kg}$ travels at a constant speed of $25 \text{ m s}^{-1}$ along a straight horizontal road. The driving force provided by the engine is $800 \text{ N}$ .
(a) State the magnitude of the resistive forces acting on the car.
[1]

(b) Calculate the power developed by the engine.
[2]

3. A ball is thrown vertically upwards with an initial velocity of $15 \text{ m s}^{-1}$ . Air resistance is negligible.
(a) Calculate the maximum height reached by the ball.
[2]

(b) Sketch the velocity-time graph for the motion of the ball from the instant it is thrown until it returns to the starting height. Label the axes with appropriate values.
[2]

4. Two trolleys, A and B, move along a smooth horizontal track. Trolley A has a mass of $2.0 \text{ kg}$ and moves with a velocity of $3.0 \text{ m s}^{-1}$ to the right. Trolley B has a mass of $1.0 \text{ kg}$ and is initially at rest. The trolleys collide and stick together.
Calculate the common velocity of the trolleys after the collision.
[3]

5. A uniform plank AB of length $4.0 \text{ m}$ and weight $200 \text{ N}$ rests on two supports. Support X is at end A, and support Y is $1.0 \text{ m}$ from end B. A boy of weight $400 \text{ N}$ stands on the plank at a distance $x$ from A.
(a) Draw a free-body diagram showing all forces acting on the plank.
[2]

(b) Determine the maximum value of $x$ such that the plank does not tip over support Y.
[3]

Section B: Data Interpretation and Problem Solving

Answer all questions in this section.

6. A student investigates the motion of a falling sphere using a data logger. The sphere is dropped from rest in a tall cylinder filled with oil. The graph below shows the variation of velocity $v$ with time $t$ .

(Imagine a graph where velocity increases rapidly at first, then the gradient decreases, eventually becoming horizontal at $v = 0.80 \text{ m s}^{-1}$ )

(a) Describe and explain the shape of the graph in terms of the forces acting on the sphere.
[3]

(b) The mass of the sphere is $0.050 \text{ kg}$ . Calculate the drag force acting on the sphere when it reaches terminal velocity.
[2]

7. A block of mass $5.0 \text{ kg}$ is pulled up a rough inclined plane by a constant force $F$ parallel to the slope. The plane is inclined at $30^\circ$ to the horizontal. The block moves at a constant speed. The frictional force acting on the block is $12 \text{ N}$ .

(a) Draw a free-body diagram for the block, labeling all forces.
[2]

(b) Calculate the magnitude of the force $F$ .
[3]

(ii) the gain in gravitational potential energy of the block.
[2]

(iii) Explain why the work done by $F$ is greater than the gain in gravitational potential energy.
[1]

8. A projectile is launched from ground level with an initial velocity of $40 \text{ m s}^{-1}$ at an angle of $30^\circ$ to the horizontal. Air resistance is negligible.

(a) Calculate the horizontal and vertical components of the initial velocity.
[2]

(b) Calculate the time of flight of the projectile.
[3]

9. Two ice skaters, P and Q, stand facing each other on frictionless ice. Skater P has a mass of $60 \text{ kg}$ and Skater Q has a mass of $80 \text{ kg}$ . They push against each other and move apart. Skater P moves with a velocity of $2.0 \text{ m s}^{-1}$ to the left.

(a) Calculate the velocity of Skater Q.
[3]

(b) Calculate the total kinetic energy of the system after they push apart.
[3]

10. A crane lifts a load of mass $500 \text{ kg}$ vertically upwards from rest with a constant acceleration of $0.50 \text{ m s}^{-2}$ .

(a) Calculate the tension in the cable during the acceleration phase.
[3]

(b) The load reaches a speed of $4.0 \text{ m s}^{-1}$ . Calculate the height gained during this acceleration.
[2]

(c) The crane then lifts the load at a constant speed of $4.0 \text{ m s}^{-1}$ . Calculate the power output of the crane motor during this constant speed phase.
[2]

11. A spring obeys Hooke's Law. When a force of $10 \text{ N}$ is applied, the extension is $5.0 \text{ cm}$ .

(a) Calculate the spring constant $k$ .
[2]

(b) Calculate the elastic potential energy stored in the spring when the extension is $5.0 \text{ cm}$ .
[2]

(c) The spring is compressed by $10 \text{ cm}$ and used to launch a ball of mass $0.10 \text{ kg}$ horizontally. Assuming all elastic potential energy is converted to kinetic energy, calculate the launch speed of the ball.
[3]

12. A car of mass $1000 \text{ kg}$ travels around a horizontal circular bend of radius $50 \text{ m}$ . The maximum frictional force between the tires and the road is $8000 \text{ N}$ .

(a) Identify the force that provides the centripetal acceleration.
[1]

(b) Calculate the maximum speed at which the car can travel around the bend without skidding.
[3]

13. A ball of mass $0.20 \text{ kg}$ is dropped from a height of $2.0 \text{ m}$ onto a hard floor. It rebounds to a height of $1.5 \text{ m}$ .

(a) Calculate the speed of the ball just before it hits the floor.
[2]

(b) Calculate the speed of the ball just after it leaves the floor.
[2]

14. A uniform ladder of weight $W$ and length $L$ rests against a smooth vertical wall and on a rough horizontal ground. The ladder makes an angle $\theta$ with the ground.

(a) Explain why the wall exerts only a horizontal force on the ladder.
[1]

(b) By taking moments about the base of the ladder, derive an expression for the horizontal force exerted by the wall in terms of $W$ and $\theta$ .
[3]

15. A rocket of mass $1000 \text{ kg}$ (including fuel) is launched vertically. The engine produces a constant thrust of $15000 \text{ N}$ . The mass of the fuel burnt is negligible for this short duration.

(a) Calculate the initial acceleration of the rocket.
[3]

(b) As the rocket rises, air resistance becomes significant. State and explain how the acceleration of the rocket changes as its speed increases.
[2]

16. A block of mass $2.0 \text{ kg}$ slides down a smooth curved track from a height of $3.0 \text{ m}$ and onto a rough horizontal surface. The coefficient of kinetic friction between the block and the horizontal surface is $0.40$ .

(a) Calculate the speed of the block at the bottom of the curved track.
[2]

(b) Calculate the distance the block travels on the horizontal surface before coming to rest.
[3]

17. Two forces, $F_1 = 10 \text{ N}$ acting horizontally to the right, and $F_2 = 10 \text{ N}$ acting at $60^\circ$ above the horizontal to the right, act on a particle.

(a) Calculate the horizontal component of the resultant force.
[2]

(b) Calculate the vertical component of the resultant force.
[1]

18. A satellite orbits the Earth in a circular orbit.

(a) State the direction of the resultant force acting on the satellite.
[1]

(b) Explain why the satellite is accelerating even though its speed is constant.
[2]

19. A student measures the acceleration of free fall $g$ using a free-fall apparatus. The time $t$ taken for a steel ball to fall a distance $h$ is recorded.

(a) State the equation relating $h$ , $t$ , and $g$ for an object falling from rest.
[1]

(b) The student plots a graph of $h$ against $t^2$ . State what physical quantity is represented by the gradient of this graph.
[1]

20. A car accelerates uniformly from rest to $20 \text{ m s}^{-1}$ in $10 \text{ s}$ . It then travels at this constant speed for $20 \text{ s}$ before decelerating uniformly to rest in $5 \text{ s}$ .

(a) Sketch the velocity-time graph for the entire journey.
[2]

(b) Calculate the total distance traveled by the car.
[3]

End of Paper

Answers

TuitionGoWhere Exam Practice (AI) - Physics H1 A-Level

PRACTICE PAPER - VERSION 5 - ANSWER KEY

Subject: Physics
Level: H1 (8867)
Total Marks: 60

Section A: Structured Questions

1. State the principle of conservation of linear momentum. [2]

Answer: In a closed system (or isolated system) [B1], the total momentum before interaction is equal to the total momentum after interaction (provided no external forces act) [B1].
Note: Accept "Total momentum remains constant if net external force is zero."

2. Car motion. [3]

(a) Resistive forces: Since speed is constant, acceleration is zero. By Newton's First Law, driving force = resistive force.
- Answer: $800 \text{ N}$ [B1]
(b) Power developed: $P = Fv$ $P = F v$
- Answer: $P = 800 \times 25 = 20,000 \text{ W}$ (or $20 \text{ kW}$ ) [M1, A1]

3. Vertical projectile. [4]

(a) Max height: Using $v^2 = u^2 + 2as$ $v^{2} = u^{2} + 2 a s$ . At max height, $v=0$ $v = 0$ . $u=15$ $u = 15$ , $a=-9.81$ $a = - 9.81$ .
- $0 = 15^2 + 2(-9.81)s$
- $s = \frac{225}{19.62} = 11.47 \text{ m}$
- Answer: $11.5 \text{ m}$ (3 s.f.) [M1, A1]
(b) Graph: Straight line with negative gradient starting at $+15$ $+ 15$ on y-axis, crossing t-axis, ending at $-15$ $- 15$ .
- Answer: Linear slope downwards [B1]; Intercepts/correct shape indicated [B1].

4. Inelastic collision. [3]

Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$ $m_{A} u_{A} + m_{B} u_{B} = (m_{A} + m_{B}) v$
- $(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0)v$
- $6.0 = 3.0v$
- $v = 2.0 \text{ m s}^{-1}$
- Answer: $2.0 \text{ m s}^{-1}$ to the right [M1, M1, A1]

5. Plank equilibrium. [5]

(a) Free-body diagram:
- Weight of plank ( $200 \text{ N}$ ) acting downwards at center ( $2.0 \text{ m}$ from A). [B1]
- Weight of boy ( $400 \text{ N}$ ) acting downwards at distance $x$ . [B1]
- Reaction at X ( $R_X$ ) upwards at A. [B1]
- Reaction at Y ( $R_Y$ ) upwards at $3.0 \text{ m}$ from A. [B1]
- (Note: Award marks for correct labels and directions)
(b) Max $x$ $x$ before tipping:
- Condition for tipping: Reaction at X becomes zero ( $R_X = 0$ ).
- Take moments about support Y.
- Clockwise moment = Anti-clockwise moment.
- Weight of plank acts at $2.0 \text{ m}$ from A. Support Y is at $3.0 \text{ m}$ from A. Distance from Y to center = $1.0 \text{ m}$ .
- Moment of plank weight about Y: $200 \times 1.0 = 200 \text{ N m}$ (Anti-clockwise).
- Moment of boy about Y: $400 \times (3.0 - x)$ (Clockwise, assuming boy is to the right of Y? No, boy is at $x$ . If $x > 3$ , moment is clockwise. If $x < 3$ , moment is anti-clockwise. Wait. To tip over Y, the boy must be to the right of Y, creating a clockwise moment that overcomes the plank's weight moment? No. The plank weight creates a moment trying to rotate it back down. The boy creates a moment trying to tip it.
- Let's check positions: A(0), Center(2), Y(3), B(4).
- Pivot at Y.
- Plank weight ( $200 \text{ N}$ ) is at $2 \text{ m}$ from A. Distance to Y = $1 \text{ m}$ to the left. Moment = $200 \times 1 = 200 \text{ Nm}$ (Counter-Clockwise).
- Boy ( $400 \text{ N}$ ) is at $x$ . To tip, boy must be to the right of Y ( $x > 3$ ). Distance to Y = $x - 3$ . Moment = $400(x - 3)$ (Clockwise).
- Equilibrium limit: $200 = 400(x - 3)$
- $0.5 = x - 3$
- $x = 3.5 \text{ m}$
- Answer: $3.5 \text{ m}$ [M1, M1, A1]

Section B: Data Interpretation and Problem Solving

6. Falling sphere in oil. [6]

(a) Shape explanation:
- Initially, velocity is zero, so drag is zero. Net force = weight. Acceleration is max ( $g$ ). [B1]
- As velocity increases, drag force increases. Net force ( $W - D$ ) decreases, so acceleration decreases. [B1]
- Eventually, drag equals weight. Net force is zero. Acceleration is zero. Velocity becomes constant (terminal velocity). [B1]
(b) Drag at terminal velocity:
- At terminal velocity, $D = W = mg$ .
- $D = 0.050 \times 9.81 = 0.4905 \text{ N}$ .
- Answer: $0.49 \text{ N}$ [M1, A1]
(c) Initial acceleration:
- At $t=0$ , $v=0$ , so Drag $= 0$ .
- $F_{net} = W$ . $ma = mg \Rightarrow a = g$ .
- Answer: $9.81 \text{ m s}^{-2}$ [B1]

7. Block on inclined plane. [10]

(a) Free-body diagram:
- Weight ( $mg$ ) vertically down. [B1]
- Normal reaction ( $N$ ) perpendicular to slope. [B1]
- Friction ( $f$ ) down the slope (opposing motion). [B1]
- Applied force ( $F$ ) up the slope. [B1]
(b) Calculate $F$ $F$ :
- Resolve forces parallel to slope. Constant speed $\Rightarrow a=0$ .
- $F = mg \sin(30^\circ) + f$
- $F = (5.0)(9.81)(0.5) + 12$
- $F = 24.525 + 12 = 36.525 \text{ N}$
- Answer: $36.5 \text{ N}$ (3 s.f.) [M1, M1, A1]
(c) Work and Energy:
- (i) Work done by $F$ $F$ : $W = F \times d = 36.525 \times 4.0 = 146.1 \text{ J}$ $W = F \times d = 36.525 \times 4.0 = 146.1 J$ .
  - Answer: $146 \text{ J}$ [M1, A1]
- (ii) Gain in GPE: $\Delta PE = mgh$ $Δ P E = m g h$ . Height $h = d \sin(30^\circ) = 4.0 \times 0.5 = 2.0 \text{ m}$ $h = d sin (3 0^{\circ}) = 4.0 \times 0.5 = 2.0 m$ .
  - $\Delta PE = 5.0 \times 9.81 \times 2.0 = 98.1 \text{ J}$ .
  - Answer: $98.1 \text{ J}$ [M1, A1]
- (iii) Explanation: Work done by $F$ is used to increase GPE AND to do work against friction (dissipated as heat). [B1]

8. Projectile motion. [7]

(a) Components:
- $u_x = 40 \cos(30^\circ) = 34.64 \text{ m s}^{-1}$
- $u_y = 40 \sin(30^\circ) = 20.0 \text{ m s}^{-1}$
- Answer: $u_x = 34.6 \text{ m s}^{-1}$ , $u_y = 20.0 \text{ m s}^{-1}$ [M1, A1]
(b) Time of flight:
- Consider vertical motion. Displacement $s_y = 0$ (returns to ground).
- $s_y = u_y t + \frac{1}{2} a_y t^2$
- $0 = 20t - 4.905t^2$
- $t(20 - 4.905t) = 0$
- $t = 0$ (start) or $t = \frac{20}{4.905} = 4.077 \text{ s}$
- Answer: $4.08 \text{ s}$ [M1, M1, A1]
(c) Horizontal range:
- $R = u_x \times t = 34.64 \times 4.077 = 141.2 \text{ m}$
- Answer: $141 \text{ m}$ [M1, A1]

9. Ice skaters. [7]

(a) Velocity of Q:
- Conservation of momentum. Initial $P = 0$ .
- $m_P v_P + m_Q v_Q = 0$
- $(60)(-2.0) + (80)(v_Q) = 0$ (Taking right as positive, P moves left)
- $-120 + 80 v_Q = 0$
- $v_Q = \frac{120}{80} = 1.5 \text{ m s}^{-1}$
- Answer: $1.5 \text{ m s}^{-1}$ to the right [M1, M1, A1]
(b) Total KE:
- $KE_P = \frac{1}{2}(60)(2.0)^2 = 120 \text{ J}$
- $KE_Q = \frac{1}{2}(80)(1.5)^2 = 90 \text{ J}$
- Total $KE = 120 + 90 = 210 \text{ J}$
- Answer: $210 \text{ J}$ [M1, M1, A1]
(c) Source of energy:
- Chemical potential energy from the skaters' muscles / Internal energy. [B1]

10. Crane lifting load. [7]

(a) Tension during acceleration:
- $T - mg = ma$
- $T = m(g + a) = 500(9.81 + 0.50) = 500(10.31)$
- $T = 5155 \text{ N}$
- Answer: $5160 \text{ N}$ (3 s.f.) [M1, M1, A1]
(b) Height gained:
- $v^2 = u^2 + 2as$
- $4.0^2 = 0 + 2(0.50)s$
- $16 = 1.0 s \Rightarrow s = 16 \text{ m}$
- Answer: $16 \text{ m}$ [M1, A1]
(c) Power at constant speed:
- At constant speed, $T = mg = 500 \times 9.81 = 4905 \text{ N}$ .
- $P = Fv = T v = 4905 \times 4.0 = 19620 \text{ W}$
- Answer: $19.6 \text{ kW}$ [M1, A1]

11. Spring energy. [7]

(a) Spring constant:
- $F = kx \Rightarrow 10 = k(0.05)$
- $k = \frac{10}{0.05} = 200 \text{ N m}^{-1}$
- Answer: $200 \text{ N m}^{-1}$ [M1, A1]
(b) Elastic PE:
- $EPE = \frac{1}{2} k x^2 = \frac{1}{2}(200)(0.05)^2$
- $EPE = 100 \times 0.0025 = 0.25 \text{ J}$
- Answer: $0.25 \text{ J}$ [M1, A1]
(c) Launch speed:
- $EPE = KE \Rightarrow 0.25 = \frac{1}{2} m v^2$
- $0.25 = \frac{1}{2}(0.10)v^2$
- $0.5 = 0.10 v^2 \Rightarrow v^2 = 5$
- $v = \sqrt{5} = 2.236 \text{ m s}^{-1}$
- Answer: $2.24 \text{ m s}^{-1}$ [M1, M1, A1]

12. Circular motion. [6]

(a) Centripetal force source:
- Friction between tires and road. [B1]
(b) Max speed:
- $F_c = \frac{mv^2}{r}$
- $F_{friction} = \frac{mv^2}{r} \Rightarrow 8000 = \frac{1000 v^2}{50}$
- $8000 = 20 v^2$
- $v^2 = 400 \Rightarrow v = 20 \text{ m s}^{-1}$
- Answer: $20 \text{ m s}^{-1}$ [M1, M1, A1]
(c) Skidding explanation:
- Required centripetal force exceeds maximum static friction. The car cannot maintain the circular path and moves in a straighter line (tangentially/outwards) relative to the curve. [B1, B1]

13. Impact momentum. [7]

(a) Speed before impact:
- $v^2 = u^2 + 2as = 0 + 2(9.81)(2.0) = 39.24$
- $v = \sqrt{39.24} = 6.264 \text{ m s}^{-1}$
- Answer: $6.26 \text{ m s}^{-1}$ [M1, A1]
(b) Speed after impact:
- Rebound height $1.5 \text{ m}$ . At top, $v=0$ .
- $0 = u_{rebound}^2 - 2(9.81)(1.5)$
- $u_{rebound} = \sqrt{29.43} = 5.425 \text{ m s}^{-1}$
- Answer: $5.43 \text{ m s}^{-1}$ [M1, A1]
(c) Change in momentum:
- Take Up as positive.
- $p_{initial} = 0.20 \times (-6.264) = -1.253 \text{ N s}$
- $p_{final} = 0.20 \times (+5.425) = +1.085 \text{ N s}$
- $\Delta p = p_f - p_i = 1.085 - (-1.253) = 2.338 \text{ N s}$
- Answer: $2.34 \text{ N s}$ (upwards) [M1, M1, A1]

14. Ladder moments. [4]

(a) Smooth wall:
- Smooth surface cannot exert friction. Therefore, the reaction force is perpendicular to the surface (horizontal). [B1]
(b) Derivation:
- Take moments about the base (point of contact with ground).
- Let $R_w$ be the horizontal force from the wall.
- Moment of $R_w$ : Force $\times$ perpendicular distance. Vertical height of contact = $L \sin \theta$ .
- Moment = $R_w (L \sin \theta)$ (Clockwise/Anti-clockwise depending on side, let's say CW).
- Moment of Weight $W$ : Acts at center ( $L/2$ ). Perpendicular distance from base = $\frac{L}{2} \cos \theta$ .
- Moment = $W (\frac{L}{2} \cos \theta)$ (Opposite direction).
- Equilibrium: $R_w L \sin \theta = W \frac{L}{2} \cos \theta$
- $R_w = \frac{W \cos \theta}{2 \sin \theta} = \frac{W}{2 \tan \theta}$
- Answer: $R_w = \frac{W}{2 \tan \theta}$ [M1, M1, A1]

15. Rocket launch. [5]

(a) Initial acceleration:
- $F_{net} = Thrust - Weight$
- $F_{net} = 15000 - (1000)(9.81) = 15000 - 9810 = 5190 \text{ N}$
- $a = \frac{F_{net}}{m} = \frac{5190}{1000} = 5.19 \text{ m s}^{-2}$
- Answer: $5.19 \text{ m s}^{-2}$ [M1, M1, A1]
(b) Effect of air resistance:
- As speed increases, air resistance (drag) increases. [B1]
- This reduces the net upward force ( $Thrust - Weight - Drag$ ), so acceleration decreases. [B1]

16. Curved track and friction. [5]

(a) Speed at bottom:
- Conservation of Energy: $PE_{top} = KE_{bottom}$
- $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$
- $v = \sqrt{2(9.81)(3.0)} = \sqrt{58.86} = 7.672 \text{ m s}^{-1}$
- Answer: $7.67 \text{ m s}^{-1}$ [M1, A1]
(b) Distance on rough surface:
- Work done by friction = Loss in KE
- $f_k d = \frac{1}{2}mv^2$
- Friction force $f_k = \mu N = \mu mg = 0.40 \times 2.0 \times 9.81 = 7.848 \text{ N}$
- $KE_{bottom} = \frac{1}{2}(2.0)(7.672)^2 = 58.86 \text{ J}$ (Matches initial PE)
- $7.848 d = 58.86$
- $d = \frac{58.86}{7.848} = 7.5 \text{ m}$
- Answer: $7.5 \text{ m}$ [M1, M1, A1]

17. Vector addition. [5]

(a) Horizontal component:
- $F_{1x} = 10 \text{ N}$
- $F_{2x} = 10 \cos(60^\circ) = 5 \text{ N}$
- $R_x = 10 + 5 = 15 \text{ N}$
- Answer: $15 \text{ N}$ [M1, A1]
(b) Vertical component:
- $F_{1y} = 0$
- $F_{2y} = 10 \sin(60^\circ) = 8.66 \text{ N}$
- $R_y = 8.66 \text{ N}$
- Answer: $8.66 \text{ N}$ [B1]
(c) Resultant magnitude:
- $R = \sqrt{R_x^2 + R_y^2} = \sqrt{15^2 + 8.66^2} = \sqrt{225 + 75} = \sqrt{300}$
- $R = 17.32 \text{ N}$
- Answer: $17.3 \text{ N}$ [M1, A1]

18. Satellite orbit. [3]

(a) Direction of force:
- Towards the center of the Earth. [B1]
(b) Acceleration explanation:
- Velocity is a vector (speed and direction). [B1]
- Although speed is constant, the direction of motion is constantly changing. Therefore, velocity is changing, which means there is acceleration. [B1]

19. Free fall experiment. [4]

(a) Equation:
- $h = \frac{1}{2}gt^2$ [B1]
(b) Gradient:
- Graph of $h$ (y) vs $t^2$ (x). Equation $y = (\frac{1}{2}g)x$ .
- Gradient = $\frac{1}{2}g$ [B1]
(c) Systematic error:
- Example: Delay in timer starting (electromagnet release time). [B1]
- Effect: Measured time $t$ is larger than actual fall time. Calculated $g$ will be smaller than actual $g$ (since $g \propto 1/t^2$ ). [B1]
- Alternative: Air resistance. Effect: $g$ calculated is lower.

20. Velocity-time graph journey. [7]

(a) Sketch:
- 0-10s: Straight line from $(0,0)$ to $(10,20)$ . [B1]
- 10-30s: Horizontal line at $v=20$ . [B1]
- 30-35s: Straight line from $(30,20)$ to $(35,0)$ . [B1] (Shape correct)
(b) Total distance:
- Area under graph.
- Area 1 (Triangle): $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$
- Area 2 (Rectangle): $20 \times 20 = 400 \text{ m}$
- Area 3 (Triangle): $\frac{1}{2} \times 5 \times 20 = 50 \text{ m}$
- Total = $100 + 400 + 50 = 550 \text{ m}$
- Answer: $550 \text{ m}$ [M1, M1, A1]
(c) Average speed:
- Avg Speed = Total Distance / Total Time
- Total Time = $10 + 20 + 5 = 35 \text{ s}$
- Avg Speed = $\frac{550}{35} = 15.71 \text{ m s}^{-1}$
- Answer: $15.7 \text{ m s}^{-1}$ [M1, A1]