Connecting our SUVAT equations to Connecting our SUVAT equations to velocity/time graphs, an alternative velocity/time graphs, an alternative view of derivationview of derivation
Learning Objectives :Learning Objectives :
1.1. To revise distance/time and speed/time To revise distance/time and speed/time graphs if necessary.graphs if necessary.
2.2. To show a connection between the To show a connection between the SUVAT equations & velocity/time graphsSUVAT equations & velocity/time graphs
3.3. To work through lots more examples of To work through lots more examples of using the SUVAT equationsusing the SUVAT equations
Book Reference : Pages 112-118Book Reference : Pages 112-118
Consider some generalised motion, starting with Consider some generalised motion, starting with an initial velocity an initial velocity uu, accelerating with acceleration , accelerating with acceleration aa for a period for a period t t ending with a final velocity ending with a final velocity vv
u
v
Vel
ocity
/ m
s-1
a
ttime / s
δv
δt Note δ means small change in
From our existing knowledge of From our existing knowledge of velocity/time graphs we know that velocity/time graphs we know that acceleration is given by the gradient of the acceleration is given by the gradient of the line:line: a =a = δδvv
δδtt
So change in velocity, So change in velocity, δδvv = = aaδδtt. .
So starting at u & after acceleration So starting at u & after acceleration aa for for tt secondssecondsv = u + at v = u + at (SUVAT 1)(SUVAT 1)
Note δ means small change in
Also from our existing knowledge of Also from our existing knowledge of velocity/time graphs we know that the velocity/time graphs we know that the distance travelled is the total area under distance travelled is the total area under the graphthe graph
u
v
Vel
ocity
/ m
s-1
a
ttime / s
v - u
Considering the areas:Considering the areas:
For the rectangle : For the rectangle : area = utarea = ut
For the triangle :For the triangle : area = ½ base x heightarea = ½ base x heightarea = ½ t (v – u)area = ½ t (v – u)
Total area = ut + ½ t (v – u)Total area = ut + ½ t (v – u)2s = 2ut + t(v – u)2s = 2ut + t(v – u)
s = s = (u + v)t(u + v)t (SUVAT 2)(SUVAT 2)
22
An alternative view:An alternative view:We are actually trying to find the area of We are actually trying to find the area of a trapezium which is:-a trapezium which is:-
Half the sum of the parallel sides x the Half the sum of the parallel sides x the distance between the parallel sidesdistance between the parallel sides
½(u + v) t½(u + v) t
s = s = (u + v)t(u + v)t (SUVAT 2)(SUVAT 2)
22
Considering the distance, (area under the Considering the distance, (area under the graph) this time without graph) this time without vv
u
v
Vel
ocity
/ m
s-1
a
ttime / s
v - u
Total area = rectangle + triangles = ut + ½(v – u) t
As before remove our dependence upon As before remove our dependence upon v v by substituting for it from SUVAT 1by substituting for it from SUVAT 1
v = u + atv = u + at
s = s = ut +ut + ½((u + at) – u) t½((u + at) – u) t
s = ut + ½ats = ut + ½at22 (SUVAT 3)(SUVAT 3)
Considering the distance, (area under the Considering the distance, (area under the graph) this time without graph) this time without tt
u
v
Vel
ocity
/ m
s-1
a
ttime / s
v - u
Total area = rectangle + triangles = ( u + v )t
2
As before remove our dependence upon As before remove our dependence upon t t by substituting for it from SUVAT 1by substituting for it from SUVAT 1
v = u + at v = u + at t = t = v – uv – u aa
s = s = ( u + v )t( u + v )t 22
s = s = (u + v) (v – u )(u + v) (v – u )2a2a
2as = uv + v2as = uv + v22 – uv – u – uv – u22
vv22 = u = u22 + 2as + 2as (SUVAT 4)(SUVAT 4)