Question Video: Determining Acceleration from a Distance–Time Graph | Nagwa Question Video: Determining Acceleration from a Distance–Time Graph | Nagwa

Question Video: Determining Acceleration from a Distance–Time Graph Science

Which of the lines shown on the distance–time graph represents the object with the greatest acceleration?

02:22

Video Transcript

Which of the lines shown on the following distance–time graph represents the object with the greatest acceleration? Is it (A) line A, (B) line B, (C) line C, or (D) the objects have equal acceleration?

The question is asking us about the accelerations of some objects. So let’s remind ourselves first of the definition of acceleration. Acceleration is defined as the rate of change of speed. In other words, the acceleration of an object is given by how much the speed of an object changes divided by the time taken for the change to occur. It is the change in speed divided by the change in time.

On the distance–time graph, the horizontal axis represents changes in time and the vertical axis represents changes in distance. Looking at the three lines on the graph, we can see that they are all straight.

Recall that on a distance–time graph, the slope of a line corresponds to the speed of the object that the line represents. And since each of these lines has a constant slope, each line represents an object moving at a constant speed. Each line has a different slope and therefore represents a different speed. Line A has a greater slope than line B, and line B has a greater slope than line C. We therefore know that the speed of A is greater than the speed of B, which is greater than the speed of C.

It is important to understand that although the speeds shown are all different, none of the speeds are changing. They’re all constant values of speed. Therefore, none of the objects are accelerating. The acceleration for all the objects is zero. This means that the correct answer to the question is option (D). All of the objects have equal acceleration, specifically zero.

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