Kinematic practice q help (1 Viewer)

[xibu34]

For c since v = s/t
then displacement= t*v right ?
so since it stops at 2 seconds wouldn't displacement be, (2)(4)=8m
in the solutions it says that the displacement is 29m, Could anyone help me see where i went wrong?

 

[yanujw]

The car stops when v=0 (i.e. the graph crosses the x-axis). The displacement at that point is the area under the curve between 0 and 8 seconds, which should give you 29m.

One more thing, the displacement at 2 seconds is actually 4m, not 8. Always look for the area under the curve, because the equation assumes constant velocity which is not true for this graph.

 

[Anounymouse352]

displacement is area under the graph
so just split into segments of triangles and rectangles

 

[xibu34]

The car stops when v=0 (i.e. the graph crosses the x-axis). The displacement at that point is the area under the curve between 0 and 8 seconds, which should give you 29m.

One more thing, the displacement at 2 seconds is actually 4m, not 8. Always look for the area under the curve, because the equation assumes constant velocity which is not true for this graph.

so would that be like this ?

 

[yanujw]

so would that be like this ?

Not quite... 'area under the curve' is used to mean area between the x-axis and the line itself.
For example, 'area under the curve' between 0 and 2 seconds is the triangle bound by (0,0) , (1,0) and (1,2) = 4, giving a displacement of 4m in this interval.

 

[xibu34]

Not quite... 'area under the curve' is used to mean area between the x-axis and the line itself.
For example, 'area under the curve' between 0 and 2 seconds is the triangle bound by (0,0) , (1,0) and (1,2) = 4, giving a displacement of 4m in this interval.

I thought so, but the solution said 29m so I assumed it would have to be a greater area.