A particle starts from rest with an acceleration of #2# #ms^-2# for #20# #s# and then comes to rest within next #20# #s# with retarding acceleration #2# #ms^-2#. How would you express the motion on a v t graph?

1 Answer
David G.
Aug 2, 2017

Please read the description below, and graph two straight lines on the v t graph: the first from #(0,0)# to #(40,20)#, the second from #(40,20)# to #(0,40)#.

Explanation:

We can calculate the maximum velocity the particle will attain:

#v=u+at=0+2xx20=40# #ms^-1#

Since the acceleration - the gradient of a v t graph - is constant, the line on the graph will begin at the origin - #v=0# #ms^-1# at #t=0# #s#. It will climb with a constant slope to a value of #v=40# #ms^-1# at #t=20# #s#, then decrease with a constant negative slope back to #v=0# #ms^-1# at #t=40# #s#.

The shape of the graph will be a triangle.

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