Motion and Time - Distance-Time Graphs

From a distance-time graph, a cyclist is seen to cover a distance of $30\ m$ in $6\ s$ at a constant speed. Calculate the speed and describe the shape of the graph.

Given: $Distance = 30\ m$, $Time = 6\ s$. Using the formula $v = \frac{d}{t}$, we get $v = \frac{30\ m}{6\ s} = 5\ m/s$.

Since the speed is constant at $5\ m/s$, the distance-time graph will be a straight line starting from the origin $(0,0)$ and passing through the point $(6, 30)$.

A car's position on a graph stays at $15\ km$ for the time interval between $t = 2\ h$ and $t = 5\ h$. What is the speed of the car during this interval?

$v = \frac{d_2 - d_1}{t_2 - t_1} = \frac{15\ km - 15\ km}{5\ h - 2\ h} = \frac{0\ km}{3\ h} = 0\ km/h$.

Because the distance does not change as time passes, the graph is a horizontal line. This indicates the car is stationary (at rest).

Observe the provided distance-time graph for a toy car. Using the data points $(2, 4)$ and $(4, 8)$, where time is in seconds and distance is in meters, calculate the speed of the car.

$Speed = \frac{d_2 - d_1}{t_2 - t_1}$ $Speed = \frac{8\ m - 4\ m}{4\ s - 2\ s}$ $Speed = \frac{4\ m}{2\ s} = 2\ m/s$

The speed is determined by finding the slope of the line. By taking two points on the line, we calculate the change in distance over the change in time.

A bus moves from Point A to Point B in $2$ hours and remains there for $2$ hours before returning. The graph shows the first $4$ hours of this journey. What is the distance of the bus from the starting point at $t = 3\ h$?

$At\ t = 2\ h,\ Distance = 40\ km$ $At\ t = 3\ h,\ Distance = 40\ km$ $At\ t = 4\ h,\ Distance = 40\ km$

Between $t = 2\ h$ and $t = 4\ h$, the graph is a horizontal line. This signifies the bus is stationary at Point B, which is $40\ km$ away from the start. Therefore, at $t = 3\ h$, its distance remains $40\ km$.