there is a barrier 50m beyond the finnishing line, find an expreSsion fo the skier speed when she is 40m beyond the finishing line?! How fast is she travelling when she is 40m beyond the finishing line! How far short of the barrier does she come to a stop?! Display an xv graph to illustrate motion?
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use v^2 = u^2 + 2as
where v is the speed after travelling a distance
u is the initial speed (30 m/s)
a is the acceleration ( - 10 m/s^2 ....note that it is negative )
and s is the distance covered.
As the question asks for an S V graph you must create a table of values. Using values of s from 0 to 50 m
I would suggest every 5 m.
And for each value calculating the appropriate value of v using the formula.
So I shouldn't tell you more than this.
Go do it!
where v is the speed after travelling a distance
u is the initial speed (30 m/s)
a is the acceleration ( - 10 m/s^2 ....note that it is negative )
and s is the distance covered.
As the question asks for an S V graph you must create a table of values. Using values of s from 0 to 50 m
I would suggest every 5 m.
And for each value calculating the appropriate value of v using the formula.
So I shouldn't tell you more than this.
Go do it!
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When people ask for more info they need to open their mailbox or I can't add anything.
And if you actually drew the graph as asked to then the answers come from that with no effort.
v=sqrt( 30^2 - 20 * 40) ( for 40 m beyond the finish. )
0= 30^2 - 20 s
so s = 30^2 / 20 = 45 m -> 5m before fence
And if you actually drew the graph as asked to then the answers come from that with no effort.
v=sqrt( 30^2 - 20 * 40) ( for 40 m beyond the finish. )
0= 30^2 - 20 s
so s = 30^2 / 20 = 45 m -> 5m before fence
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