A train of mass 5.6 × 10⁵ kg is at rest in a station.
At time t = 0 s, a resultant force acts on the train and it starts to accelerate forwards.
The graph is the distance-time graph for the train for the first 120 s.

(a) (i) Use the distance-time graph to determine the average speed of the train during the 120 s.

\(\) \text{average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{4800 \text{ m}}{120 \text{ s}} = \boxed{40 \text{ m/s}} \(\)

(ii) Use the distance-time graph to determine the speed of the train at time t = 100 s.

The speed at a given time on a distance-time graph is represented by the slope of the graph at that time. To find the speed at *t* = 100 s, we can draw a tangent to the curve at that point and calculate its gradient. From the graph, the tangent at *t* = 100 s passes through approximately (80 s, 2400 m) and (120 s, 4800 m). Therefore, the speed at *t* = 100 s is: \(\) \text{speed} = \frac{\text{change in distance}}{\text{change in time}} = \frac{4800 \text{ m} - 2400 \text{ m}}{120 \text{ s} - 80 \text{ s}} = \boxed{60 \text{ m/s}} \(\)

(iii) Describe how the acceleration of the train at time t = 100 s differs from the acceleration at time t = 20 s.

The acceleration of an object is represented by the **curvature** of its distance-time graph. * A **concave upwards** curve indicates **increasing acceleration.** * A **straight line** indicates **constant acceleration.** * A **concave downwards** curve indicates **decreasing acceleration.** At *t* = 20 s, the graph is concave upwards, indicating that the train is accelerating. At *t* = 100 s, the graph is a straight line, meaning the train is moving at a constant speed, and therefore its acceleration is **zero**.

**(b) (i)** The initial acceleration of the train is 0.75 m/s².
Calculate the resultant force that acts on the train at this time.

Using Newton's Second Law of Motion: \(\) F = ma \(\) where: * *F* is the resultant force (N) * *m* is the mass (kg) * *a* is the acceleration (m/s²) Therefore, the resultant force is: \(\) F = (5.6 \times 10^5 \text{ kg}) \times (0.75 \text{ m/s}^2) = \boxed{4.2 \times 10^5 \text{ N}} \(\)

(ii) At time t = 120 s, the train begins to decelerate.
State what is meant by deceleration.

Deceleration is the rate of decrease of speed, or simply **negative acceleration**.

A student determines the speed of three cars on a straight road.
The student measured the time for the cars to travel 50 m.
The table shows the measurements.

(a) (i) Without calculation, identify the fastest car and the slowest car.
Complete the table.

The car that takes the shortest time to travel the same distance is the fastest. Conversely, the car that takes the longest time is the slowest. Therefore:

(ii) Calculate the speed of car B.

\(\) \text{speed} = \frac{\text{distance}}{\text{time}} = \frac{50 \text{ m}}{4.0 \text{ s}} = \boxed{12.5 \text{ m/s}} \(\)

**(b) (i)** Estimate the time, in minutes, for car C to travel 5000 m.

First, find the speed of car C: \(\) \text{speed} = \frac{\text{distance}}{\text{time}} = \frac{50 \text{ m}}{3.6 \text{ s}} = 13.89 \text{ m/s} \(\) Then calculate the time to travel 5000 m: \(\) \text{time} = \frac{\text{distance}}{\text{speed}} = \frac{5000 \text{ m}}{13.89 \text{ m/s}} = 360 \text{ s} \(\) Finally, convert the time to minutes: \(\) \text{time in minutes} = \frac{360 \text{ s}}{60 \text{ s/min}} = \boxed{6 \text{ minutes}} \(\)

**(ii)** Explain why your answer in (b)(i) may not be the same as the actual time taken for the car to travel 5000 m.

The calculation assumes that the car travels at a constant speed. In reality, the car may accelerate, decelerate, or stop during the 5000 m journey, affecting the actual time taken.

A woman drives a car from town A to town B. She stops at a garage during her journey.

The distance-time graph for the journey is shown on the graph.

**(a) (i)** Determine the total time for the whole journey.

The total time for the journey is the time at which the graph ends. From the graph, the total time is **2.5 h**.

**(ii)** Determine the time for which the car is not moving.

The car is not moving when the distance stays constant over time, which is represented by a horizontal line on the graph. From the graph, the car is not moving between 0.75 h and 1.25 h, so the time for which it is not moving is 1.25 h - 0.75 h = **0.5 h**.

**(iii)** Determine the distance between town A and town B.

The distance between the towns is the final distance reached by the car. From the graph, the distance is **100 km**.

**(iv)** Calculate the average speed of the car between 0 and 0.75 h.

\(\) \text{average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{75 \text{ km}}{0.75 \text{ h}} = \boxed{100 \text{ km/h}} \(\)

**(b)** The speed of the car before stopping at the garage is different from its speed after stopping at the garage.
Describe this difference in speed and explain how the graph shows it.

The **slope** of a distance-time graph represents **speed.** * A **steeper** slope indicates a **higher speed.** * A **shallower** slope indicates a **lower speed.** Before stopping at the garage, the graph is steeper, indicating a higher speed. After stopping at the garage, the graph is shallower, indicating a lower speed. Therefore, the car was traveling **faster before** stopping at the garage than **after**.

An aeroplane of mass 2.5 × 10⁵ kg lands with a speed of 62 m/s, on a horizontal runway at time t = 0. The aeroplane decelerates uniformly as it travels along the runway in a straight line until it reaches a speed of 6.0 m/s at t = 35 s.

**(a)** Calculate the deceleration of the aeroplane in the 35 s after it lands.

\(\) \text{deceleration} = \frac{\text{change in speed}}{\text{time}} = \frac{6.0 \text{ m/s} - 62 \text{ m/s}}{35 \text{ s}} = \boxed{-1.6 \text{ m/s}^2} \(\)

**(b)** Calculate the resultant force acting on the aeroplane as it decelerates.

Using Newton's Second Law of Motion: \(\) F = ma \(\) where: * *F* is the resultant force (N) * *m* is the mass (kg) * *a* is the acceleration (m/s²) Therefore, the resultant force is: \(\) F = (2.5 \times 10^5 \text{ kg}) \times (-1.6 \text{ m/s}^2) = \boxed{-4.0 \times 10^5 \text{ N}} \(\) The negative sign indicates that the force is acting in the opposite direction to the motion of the aeroplane.

**(c)** Calculate the momentum of the aeroplane when its speed is 6.0 m/s.

\(\) \text{momentum} = \text{mass} \times \text{velocity} = (2.5 \times 10^5 \text{ kg}) \times (6.0 \text{ m/s}) = \boxed{1.5 \times 10^6 \text{ kg m/s}} \(\)