Question

Rank the time intervals in order of increasing displacement, from largest negative to largest positive.

A bicycle is moving along a straight line. The graph shows its position from the starting point as a function of time. Consider the 1 s time intervals 0–1 s, 1–2 s, and so on.

Answer 1

The displacement of object is equal to distance travelled between final position and initial position in the corresponding time intervals shown in the given graph.

 

The position of an object at time t1 is x1 and at time t2 be x2. Thus, the displacement of object is,

x12 = x2 - x1/t2 – t1

 

The graph between displacement x and time t is shown in the following figure.

 

From the given graph, the displacement of object in the interval 0 s to 2 s is,

xAB = 20m – 0m

       = 20m

 

From the given graph, the displacement of object in the interval 2 s to 3 s is,

xBC = 20m – 20m

       = 0m

 

From the given graph, the displacement of object in the interval 3 s to 4 s is,

xCD = 30m – 20m

        = 10m

 

From the given graph, the displacement of object in the interval 5 s to 4 s is,

xDE = 0m – 30m

       = -30m

 

From the given graph, the displacement of object in the interval 5 s to 6 s is,

xEF = 20m - 0m

       = 20m

 

Thus, the displacement of object in order of increasing displacement is

xDE < xBC < xCD < xAB = xEF.

 

Therefore, the rank of the interval in order of the speed from smallest to largest is 4s to 5s, 2s to 3s, 3s to 4s, 0 to 2s, and 5s to 6s.

Therefore, the rank of the interval in order of the speed from smallest to largest is 4s to 5s, 2s to 3s, 3s to 4s, 0 to 2s, and 5s to 6s.