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M1 help needed urgently

7. Two ramblers, Alison and Bill, are out walking. At midday, Alison is at the fixed origin O,and Bill is at the point with position vector (5i + 12j) km relative to O, where i and j are perpendicular, horizontal unit vectors.
They are both walking with constant velocity Alison at (2i + 5j) km h^-1, and Bill at a speed
of 2sqrt10 km h^-1 in a direction parallel to the vector (3i + j).
(b) Show that the velocity of Bill is (6i + 2j) km h^-1. (3 marks)
(c) Show that, at time t hours after midday, the position vector of Bill relative to Alison is
[(4t – 5)i + (12 – 3t)j] km. (5 marks)
(d) Show that the distance, d km, between the two ramblers is given by
d 2 = 25t 2 112t + 169. (2 marks)
(e) Using your answer to part (d), find the length of time to the nearest minute for which
the distance between the Alison and Bill is less than 11 km.
(5 marks)
(edited 13 years ago)
Reply 1
Have you tried any of these yourself yet? Here's some general advice.

(b) First get the direction of the vector as an angle from the x axis. Then use trigonometry to get the horizontal and vertical component of the velocity.

(c) Calculate the position vectors of Bill and Alice at time t then subtract one from the other.

(d) Use pythagorus

(e) Solve 25t2112t+169=11225t^2-112t+169=11^2 to get two values of t for when the distance is exactly 11km. Then subtract one value of t from the other.

This is the sort of question where if you can't do one part then later parts are possible so if you're struggling with something early on then move on to do later parts of the question.
Reply 2
Original post by sammy3000
7. Two ramblers, Alison and Bill, are out walking. At midday, Alison is at the fixed origin O,and Bill is at the point with position vector (5i + 12j) km relative to O, where i and j are perpendicular, horizontal unit vectors.
They are both walking with constant velocity Alison at (2i + 5j) km h^-1, and Bill at a speed
of 2sqrt10 km h^-1 in a direction parallel to the vector (3i + j).
(b) Show that the velocity of Bill is (6i + 2j) km h^-1. (3 marks)
(c) Show that, at time t hours after midday, the position vector of Bill relative to Alison is
[(4t – 5)i + (12 – 3t)j] km. (5 marks)
(d) Show that the distance, d km, between the two ramblers is given by
d 2 = 25t 2 112t + 169. (2 marks)
(e) Using your answer to part (d), find the length of time to the nearest minute for which
the distance between the Alison and Bill is less than 11 km.
(5 marks)


b) The unit vector for movemnet of Bill is
eB=vv=(310i+110j)\displaystyle \vec {e_B}=\frac{\vec {v}}{|\vec{v}|}=\left( \frac{3}{\sqrt{10}}i+\frac{1}{\sqrt{10}}j \right)
Multiplying this with the 2102\sqrt{10} velocity you get the velocity vector of Bill
c) the position vector of Bill after t hours will be
rB=r0B+tvB=(5+6t)i+(12+2t)j\displaystyle \vec {r'_B}=\vec{r_{0B}}+t\vec{v_B}=(5+6t)i+(12+2t)j
For Alison
rA=r0A+tvA=(2ti+5tj)\displaystyle \vec{r'_A}=\vec{r_{0A}}+t\vec{v_A}=(2ti+5tj)
WIth a substraction of rBrA\displaystyle \vec {r'_B}-\vec{r'_A}
you get the position of Bill relative to A which is [(4t+5)i+(12-3t)j]
because Bill starting position was (5,12) and is not (-5,12)
d) For the distance between them calculate the absolute value of
rBrA=(xBxA)2+(yByA)2\displaystyle |\vec {r'_B}-\vec{r'_A}|=(x_B-x_A)^2+(y_B-y_A)^2
where x is the coordinates for i, and y is the coordinates for j)
e) Solve the equation
(edited 13 years ago)

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