A student travels from home to school. She first walks 400 m to a bus stop, taking 300 s. She then waits at the bus stop for 100 s before the bus arrives. Finally, she travels 4500 m on the bus to school, taking 300 s. (a) Calculate the total distance she travels and the total t…

Answer: ((a)) Total distance = 4900 m, total time = 700 s, average speed = 7.0 m/s ((b)) The graph would be a horizontal (flat) straight line, because her distance from home stays the same (she is not moving) while time passes. ((c)) The bus is accelerating, because the gradient of the distance-time graph is getting steeper over this time, which means its speed is increasing.. Part (a): Average speed for the whole journey First, find the total distance travelled by adding every stage of the journey (the waiting time adds no extra distance): total distance = 400 m + 4500 m = 4900 m Next, find the total time taken…

IGCSE Physics 0625: Average Speed and Interpreting a Distance-Time Graph

A student travels from home to school. She first walks $400\text{ m}$ to a bus stop, taking $300\text{ s}$ . She then waits at the bus stop for $100\text{ s}$ before the bus arrives. Finally, she travels $4500\text{ m}$ on the bus to school, taking $300\text{ s}$ .

(a) Calculate the total distance she travels and the total time taken for the whole journey (including the time spent waiting), and use these to calculate her average speed for the whole journey. [3]

(b) On a distance–time graph for this journey, state what feature of the graph would be seen during the $100\text{ s}$ she spends waiting at the bus stop, and explain why the graph has this feature. [2]

(c) As the bus pulls away from the stop, the distance–time graph for this part of the journey is a curve that gets steeper and steeper for the first $10\text{ s}$ , before becoming a straight line. State whether the bus is accelerating or decelerating during this first $10\text{ s}$ , and explain how you can tell this from the shape of the graph. [2]

Part (a): Average speed for the whole journey

First, find the total distance travelled by adding every stage of the journey (the waiting time adds no extra distance):

$\text{total distance} = 400\text{ m} + 4500\text{ m} = 4900\text{ m}$

Next, find the total time taken by adding every stage, including the waiting time:

$\text{total time} = 300\text{ s} + 100\text{ s} + 300\text{ s} = 700\text{ s}$

Average speed uses the total distance and total time for the whole journey:

$\text{average speed} = \frac{\text{total distance travelled}}{\text{total time taken}} = \frac{4900\text{ m}}{700\text{ s}}$

$\text{average speed} = 7.0\text{ m/s}$

Part (b): The graph during the waiting time

While the student waits at the bus stop, her distance from home is not increasing. It stays constant. On a distance–time graph, this is shown as a horizontal (flat) straight line. The gradient of this line is zero, and since the gradient of a distance–time graph represents speed, a gradient of zero means her speed is zero: she is at rest.

Part (c): Interpreting the curving section

As the bus pulls away, the graph is a curve, not a straight line, so the bus’s speed is not constant during this time. Because the curve gets steeper and steeper, its gradient, and therefore the bus’s speed, is increasing throughout this section. An increasing speed means the bus is accelerating.

Final answers

(a) Total distance $=$ $4900\text{ m}$ , total time $=$ $700\text{ s}$ , average speed $=$ $7.0\text{ m/s}$
(b) The graph is a horizontal line during the wait, showing she is at rest (zero speed)
(c) The bus is accelerating, because the graph’s gradient is increasing over this time