Question 1: A velocity-time graph shows how an object's velocity changes over time. The gradient of the graph represents the object's acceleration, and the area under the graph corresponds to the displacement.

Answer: A velocity-time graph illustrates velocity variation over time; gradient indicates acceleration, and area under the curve equals displacement.

Question 1: What does a straight horizontal line on a velocity-time graph indicate about the motion of an object?

Answer: It indicates the object is moving at a constant velocity, with zero acceleration.

Question 2: Explain what a positive gradient on a velocity-time graph signifies about the object's motion.

Answer: A positive gradient indicates the object is accelerating, with velocity increasing over time.

Question 1: An object accelerates uniformly from a velocity of 10 m/s to 30 m/s over 5 seconds. Calculate its acceleration in m/s².

Answer: Acceleration = (Final velocity - Initial velocity) / Time = (30 - 10) / 5 = 4 m/s²

Question 2: A car travels with a constant velocity of 20 m/s for 8 seconds. What is the total displacement during this period?

Answer: Displacement = Velocity × Time = 20 × 8 = 160 metres

Question 1: Describe a simple experiment to measure the acceleration of a toy car using a velocity-time graph. Include the variables involved and safety precautions.

Answer: Set up a ramp for the toy car to roll down. Use a motion sensor or stop-clock to record the velocity at different times, plotting these on a velocity-time graph. Variables include initial velocity, final velocity, and time. Ensure the ramp is stable, the area is clear, and the car's path is unobstructed for safety.

Question 2: Identify two limitations of interpreting velocity-time graphs in real-world scenarios.

Answer: They assume uniform acceleration which may not be true in real-world conditions, and measurements can be affected by sensor inaccuracies or timing errors.

Question 1: A cyclist accelerates from 0 m/s to 10 m/s over 5 seconds, then maintains this velocity for another 10 seconds. Calculate the total displacement during the entire 15 seconds.

Answer: Displacement during acceleration = (0 + 10)/2 × 5 = 25 metres; during constant velocity = 10 × 10 = 100 metres; total = 125 metres.

Question 2: Explain how a velocity-time graph can be used in the transportation industry to improve vehicle efficiency.

Answer: Velocity-time graphs help analyse vehicle acceleration patterns, identify periods of inefficient speeding or idling, and optimise driving strategies for fuel efficiency and reduced emissions.

Question 1: A vehicle accelerates uniformly from 5 m/s to 25 m/s over 8 seconds. Draw a velocity-time graph for this motion and calculate the acceleration. (Note: You are only required to describe the graph, not draw it).

Answer: The graph is a straight line sloping upward from 5 m/s at 0 seconds to 25 m/s at 8 seconds. The gradient = (25 - 5) / 8 = 2.5 m/s², indicating uniform acceleration.