MATHS MECHANICS HELP PLEASE!!!!

[B333]

Can you please do question b and explain how you did it. I looked at the mark scheme but it makes no sense whatsoever.

Two dogs, Fido and Growler, are playing in a field. Fido is moving in a straight line so that at time t his position vector relative to a fixed origin, O, is given by [(2t – 3)i + tj] metres. Growler is stationary at the point with position vector (2i + 5j) metres, where i and j are horizontal perpendicular unit vectors. (a) Find the displacement vector of Fido from Growler in terms of t. (2 marks)
(b) Find the value of t for which the two dogs are closest. (6 marks)
(c) Find the minimum distance between the two dogs.o dogs.

 

[Herald Grove]

Is the answer from the MS 1s or 2.5s? ... for part b

 

[Herald Grove]

Never seen such a method Hamzeh S!

 

[bloom princeton]

Can you please do question b and explain how you did it. I looked at the mark scheme but it makes no sense whatsoever.

Two dogs, Fido and Growler, are playing in a field. Fido is moving in a straight line so that at time t his position vector relative to a fixed origin, O, is given by [(2t – 3)i + tj] metres. Growler is stationary at the point with position vector (2i + 5j) metres, where i and j are horizontal perpendicular unit vectors. (a) Find the displacement vector of Fido from Growler in terms of t. (2 marks)
(b) Find the value of t for which the two dogs are closest. (6 marks)
(c) Find the minimum distance between the two dogs.o dogs.

okay buddy here it is..it took me A LOT of time to figure it out ill try to explain

a) the question basically asks the displacement vector GF so:
{(2t-3)i +t j} - (2i+5j) = (2t-5)i + (t-5)j

b) u need to find the time when the two dogs are the closest so whenever u spot something like closest..biggest..it should ring that aha i need to complete the square of an equation and as you have the displacemnt vector u know that distance is the magnitude of diplacement therefore use phythagoras theorm like so:
d^2= (2t-5)^2 + (t-5)^2
d^2= 5t^2 -30t +50 (take 5 as a common factor)
d^2= 5(t^2-6t+10) now complete the square
d^2=5{(t-3)^2 +1}
hence the time when they two dogs are closest is when t=3

c)Now that you have found the time when the two dogs are closest,u get the idea that 'hey!check it out closest distance=minimum distance!'

so just substitute value of t into the displacement vector GF
and you get (i-2j)
now all thats left is to use phaythagoras theorm to calculate the distance
like we did earlier
d^2= {(1)^2} + (-2)^2
d^2=5
therefore d=square root 5 which is equal to 2.236 =2.24m (3s.f)


IT TOOK ME A WHOLLLLE DAY AND WHEN I CHECKED THE MARKSKCHEME I WAS LIKE AHHHHHHHHHH...NOW I GET IT
SO DO YOU GET IT,IF YOU HAVE ANY DOUBT FEEL FREE TO ASK..THOUGH I MIGHT REPLY A BIT LATE

 

[Hamzeh S]

Never seen such a method Hamzeh S!

well there are multiple ways to do M1 questions this is the one my teacher taught me depends on C1 quadratic graphs vertix which is also the minimum point

 

[B333]

d^2=5{(t-3)^2 +1}
how did you complete the square wouldn't you get -1 and you can't get the square root of a negative number

 

[bloom princeton]

d^2=5{(t-3)^2 +1}
how did you complete the square wouldn't you get -1 and you can't get the square root of a negative number

no basically you cant solve it this is just like finding the minimum point or maximum point in a quadratcic curve remember what you there
you complete the square of the quadratic but you cannt say y=0,so u say the minmium point is(opposite sign of the x,same value of y) and here we are only concerned about the x value becoz(t-3) so t=3, u cannot complete the square to bget two solutions becoz u do not know what id the value of d^2 .

 

[kmkm20092008]

which exam is this question in?

 

[B333]

which exam is this question in?

it is from a solomon paper

no basically you cant solve it this is just like finding the minimum point or maximum point in a quadratcic curve remember what you there
you complete the square of the quadratic but you cannt say y=0,so u say the minmium point is(opposite sign of the x,same value of y) and here we are only concerned about the x value becoz(t-3) so t=3, u cannot complete the square to bget two solutions becoz u do not know what id the value of d^2 .

thank you so much, you're a life saver

 

[ISTIAQUE]

okay buddy here it is..it took me A LOT of time to figure it out ill try to explain

a) the question basically asks the displacement vector GF so:
{(2t-3)i +t j} - (2i+5j) = (2t-5)i + (t-5)j

b) u need to find the time when the two dogs are the closest so whenever u spot something like closest..biggest..it should ring that aha i need to complete the square of an equation and as you have the displacemnt vector u know that distance is the magnitude of diplacement therefore use phythagoras theorm like so:
d^2= (2t-5)^2 + (t-5)^2
d^2= 5t^2 -30t +50 (take 5 as a common factor)
d^2= 5(t^2-6t+10) now complete the square
d^2=5{(t-3)^2 +1}
hence the time when they two dogs are closest is when t=3

c)Now that you have found the time when the two dogs are closest,u get the idea that 'hey!check it out closest distance=minimum distance!'

so just substitute value of t into the displacement vector GF
and you get (i-2j)
now all thats left is to use phaythagoras theorm to calculate the distance
like we did earlier
d^2= {(1)^2} + (-2)^2
d^2=5
therefore d=square root 5 which is equal to 2.236 =2.24m (3s.f)


IT TOOK ME A WHOLLLLE DAY AND WHEN I CHECKED THE MARKSKCHEME I WAS LIKE AHHHHHHHHHH...NOW I GET IT
SO DO YOU GET IT,IF YOU HAVE ANY DOUBT FEEL FREE TO ASK..THOUGH I MIGHT REPLY A BIT LATE

man u ar genius!!

 

[bloom princeton]

man u ar genius!!

not really i couldnt do the moments question in the exam...