After having learned **Speed and Velocity (Part 3)** and **Acceleration (Part 4)**, **SPM Form 4 students** would be able to apply these concept into this current post of **Graphs of Linear Motion (I) SPM Physics Form 4 Notes**. Students need to understand the basic dynamic concept to interpret graphs. As Pure Physics students would be asked to interpret graphs in **SPM Physics examination**. **Berry Berry Easy **would like to present you another post on Linear Motion by our Universiti Sains Malaysia (USM) graduated contributor, See Chee Keong. He will provide interpretation of displacement versus time graphs for our **Form 4 and Form 5 Physics students**.

**SPM Physics Form 4 Notes – Force and Motion (Part 5)**

**Graphs of Linear Motion (I)**

Graph is an important tool in physics, it can be used to record the behaviour of an object in certain time interval. As a physics student, by analysing the graph, student can obtain the displacement, velocity and acceleration of the object at any particular time within the time interval. If you are a Physics student, you need to really familiarise with graphs and grab this chance to master it.

In SPM level, students always deal with the graph of motion. Therefore, in this section, I will discuss about the graphs of linear motion in detail, hope that everyone can grasp something here.

**Some tips about the graph.**

In dynamics, we usually deal with the **area** under the graph and the **gradient **of the graph. How we determine the gradient of a curve? Since some graph of motion present in the curve shape.

The gradient of a point in a curve can be determined by draw a straight line at that point (tangent) and calculate the gradient of that line. We call it the instantaneous velocity or instantaneous acceleration in corresponding graph.

In the graph above, we can obtain the gradient *m1* and *m2* by draw the line on corresponding point. By looking on the steepness of the line, easily we know that the *m1* is larger than *m2*. Since the line for *m1* is steeper than the line for *m2*. From this information we can conclude that the gradient of this graph decrease.

**Graph of displacement versus time.**

In previous section, we learned *v* = *s*/*t.* In the graph of *s* versus *t*, the gradient of the graph equal to *s*/*t*. Therefore what we can conclude is that the gradient of the graph *s* versus *t* is equal to the velocity.

In the graph above shows a linear graph, the gradient is a fixed value. From graph, obviously the object has a constant velocity.

In the graph above shows a curve, the steepness of the curve reduce. Hence we deduce that the velocity of the object decrease since the gradient decrease.

The steepness of the curve in the graph above increases, therefore the velocity keep increases as the time pass through.