Question

An object is moving along a straight line with a uniform speed of 10 m/s. Plot a graph showing distance versus time from t = 0 to t = 10 s.

Answer 1

Hint: To plot a graph of distance versus time, we need to calculate distance for t = 0 to t = 10 s. Distance is given by the formula:
\[d=s\times t\]
Where s is speed and t is time.
We will calculate distance for 10 different values of t and since speed is uniform, the value of speed will remain 10 m/s.
After calculating the values, we can plot a graph taking time on the x-axis and distance on y-axis.

Complete step by step answer:
For t = 0,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 0 \\
 & \Rightarrow d=0 \\
\end{align}\]
Therefore, when t = 0, d = 0.
For t = 1,
\[\begin{align}
 & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 1 \\
 & \therefore d=10\text{ m} \\
\end{align}\]
Therefore, when t = 1 s, d = 10 m.
For t = 2,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 2 \\
 & \therefore d=20\text{ m} \\
\end{align}\]
Therefore, when t = 2 s, d = 20 m.
For t = 3,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 3 \\
 & \therefore d=30\text{ m} \\
\end{align}\]
Therefore, when t = 3 s, d = 30 m.
For t = 4,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 4 \\
 & \therefore d=40\text{ m} \\
\end{align}\]
Therefore, when t = 4 s, d = 40 m.
For t = 5,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 5 \\
 & \Rightarrow d=50\text{ m} \\
\end{align}\]
Therefore, when t = 5 s, d = 50 m.
For t = 6,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 6 \\
 & \Rightarrow d=60\text{ m} \\
\end{align}\]
Therefore, when t = 6 s, d = 60 m.
For t = 7,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 7 \\
 & \Rightarrow d=70\text{ m} \\
\end{align}\]
Therefore, when t = 7 s, d = 70 m.
For t = 8,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 8 \\
 & \therefore d=80\text{ m} \\
\end{align}\]
Therefore, when t = 8 s, d = 80 m.
For t = 9,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 9 \\
 & \therefore d=90\text{ m} \\
\end{align}\]
Therefore, when t = 9 s, d = 90 m.
For t = 10,
\[\begin{align}
  & \text{ }d=s\times t \\
 & \Rightarrow d=10\times 10 \\
 & \therefore d=100\text{ m} \\
\end{align}\]
Therefore, when t = 10 s, d = 100 m.
Taking time on x-axis and distance on y-axis, we can obtain a graph as follows:

Note:
In the formula for distance, that is \[d=s\times t\] distance and time are directly proportional values, that is as distance increases, time also increases and vice versa. This can be verified from the graph.
We can plot the graph by taking any number of values of t. The nature of the graph will not change.