At 6 meters every second, it will clearly take less than a second to move 2.4 meters.
The ratio of displacement to displacement in a second is 2.4/ 6 = .4, so the time interval is .4 second.
Generalized Response: We can obviously rearrange
`ds = vAve `dt
into the form
`dt = `ds / vAve.
This form represents the ratio `ds / vAve of the displacement moved to the displacement moved in a second.
If, as in the present problem, the numerical value of `ds is less than that of v, this means that we have moved less displacement than would be moved in a second. So the time interval `dt is less than a second.
The smaller the ratio `ds / vAve, the less the time required.
We can reason out the result, as above. As an alternative, the figure below shows the diagram in the form of a 'relationship triangle' between `ds, `dt and vAve, showing how we can use the known quantities `ds and vAve to find the time interval `dt.
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