a maths help
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need help on 1 qtn....
s = 2t^2 - 4t +9
find the total distance travelled by the particle in the period t=0 to t=3
ds/dt = 4t- 4
then from there i not sure how to do already cause my ans wrong. pls guide me. ty -
sketch the graph of 4t - 4Originally posted by Y_Shun:need help on 1 qtn....
s = 2t^2 - 4t +9
find the total distance travelled by the particle in the period t=0 to t=3
ds/dt = 4t- 4
then from there i not sure how to do already cause my ans wrong. pls guide me. ty
then find the area under t=0 and t=3
that is the distance travelled -
integrate s wrt t. done.Originally posted by Y_Shun:need help on 1 qtn....
s = 2t^2 - 4t +9
find the total distance travelled by the particle in the period t=0 to t=3
ds/dt = 4t- 4
then from there i not sure how to do already cause my ans wrong. pls guide me. ty -
remember you cannot directly multiply the t=3 and t=0 then you minus off because that is the displacement not the distance travelled.
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then you get back the S which is already givenOriginally posted by hisoka:integrate s wrt t. done.
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ds/dt = 0 (when particle is not moving or at making u-turn point)Originally posted by Y_Shun:need help on 1 qtn....
s = 2t^2 - 4t +9
find the total distance travelled by the particle in the period t=0 to t=3
ds/dt = 4t- 4
then from there i not sure how to do already cause my ans wrong. pls guide me. ty
4t - 4 = 0
t = 1
t=0, s=9
t=1, s=7
t=3, s=15
total distance travelled
(9-7) + (15-7)
= 2 + 8
= 10 -
okOriginally posted by Y_Shun:need help on 1 qtn....
s = 2t^2 - 4t +9
find the total distance travelled by the particle in the period t=0 to t=3
ds/dt = 4t- 4
then from there i not sure how to do already cause my ans wrong. pls guide me. ty
u ds/dt
u get 4t-4 right
then this 4t-4 is the velocity. equate it to 0
t=1
at t=1, the particle changes direction
sub t=1
s= 9
sub t= 1 into S
s = 7
sub t=3 into S
s=15
dist = 9-7+ ( 15-7)
= 10 -
this is one methodOriginally posted by FireIce:ds/dt = 0 (when particle is not moving or at making u-turn point)
4t - 4 = 0
t = 1
t=0, s=9
t=1, s=7
t=3, s=15
total distance travelled
(9-7) + (15-7)
= 2 + 8
= 10
is the area under graph another method? -
FI's answer correct!Originally posted by FireIce:ds/dt = 0 (when particle is not moving or at making u-turn point)
4t - 4 = 0
t = 1
t=0, s=9
t=1, s=7
t=3, s=15
total distance travelled
(9-7) + (15-7)
= 2 + 8
= 10
but
how come must 9-7 ar? -
no...Originally posted by Diehard_89:this is one method
is the area under graph another method? -
cos the particle started at the point when it's 9mOriginally posted by Y_Shun:FI's answer correct!
but
how come must 9-7 ar?
it travelled backward (to 7m point) then make u-turn (when ds/dt = 0), then forward to 15m (when t=3) -
no?Originally posted by AMEN567889:no...
I tried already got the answer leh
area under graph which is velocity against time leh
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kaoz...u fail ur A maths izit. lolOriginally posted by Y_Shun:FI's answer correct!
but
how come must 9-7 ar?
9 is the displacement of the particle at which time = 0
then when time starts.
it moves to the left assuming ur taking < -
then i dun know le...Originally posted by Diehard_89:no?
I tried already got the answer leh
area under graph which is velocity against time leh
forget liao...
i took my O's last year... =/
now my mind is being crapmed with A level stuff... -
just learnt this topic...but still very blur abt it...got prob understanding the qtn =x
btw Diehard_89, whose your maths teacher ar? -
this is quite simple differentiation
just rber to check for the turning points
you can plot linear and mark all the points on a straight line.
no need graph -
Area under graph does give u the answer. I'm not sure whether it is allowed, although I think it should be unless the question specifically states that differentiation and integration must be used.Originally posted by Diehard_89:this is one method
is the area under graph another method?
If you use the graph method:
Just take note that the intercept of the x-axis (time) is the time for the turning point of the particle. It is the point at which the velocity is zero. The area after this point will be a negative if you do normal integration because the particle is now travelling backwards. However, we are concerned with distance here, and hence u will need to take the positive value, and add it to the area above the x-axis before this intercept.