Help Solve Part C of Vector Question for Cathy

[CathyLou]

Hi.

I would really appreciate it if someone could please explain to me part c of this M1 level vector question as I am really stuck.

Two trains A and B leave the same station, O, at 10 a.m. and travel along straight horizontal tracks. A travels with constant speed 80 Km/h due east and B travels with constant speed 52 km/h in the direction (5i + 12j) where i and j are unit vectors due east and due north respectively.

(a) Show that the velocity of B is (20i + 48j) km/h.

I got this part ok.

(b) Find the displacement vector of B from A at 10:15 a.m.

I got this as (15i + 12j).

Given that the trains are 23 km apart t minutes after 10 a.m.

(c) find the value of t correct the nearest whole number.

Thank you.

Cathy

 

[learningphysics]

Shouldn't part b) be 15i - 12j ?

For part c) you need the displacement vector from B to A (or A to B) in terms of t. Then set the magnitude of that vector equal to 23km and solve for t.

 

[CathyLou]

Thanks for your help.

I got 15 hours.

Is this correct?

Cathy

 

[learningphysics]

Thanks for your help.

I got 15 hours.

Is this correct?

Cathy

No. Can you show your calculations?

 

[CathyLou]

Ok.

I got that the position vector of A in terms of t is 80ti and that the position vector of B in terms of t = (20ti + 48tj).

I then subtracted B from A and got (60ti - 48tj).

23^2 = 529 and so the square root of ((60t)^2 - (48t)^2)) = 529

1296t^2 = 279841

t = 14.69444444 hours or 15 hours to the nearest whole number.

Cathy

 

[learningphysics]

23^2 = 529 and so the square root of ((60t)^2 - (48t)^2)) = 529

1296t^2 = 279841

It should be $$\sqrt{{60t}^2 + {(-48t)}^2} = 23$$

Remember that when getting the magnitude of the vector, to take the entire component... ie 60ti -48tj = 60ti + (-48t)j... so you need to take (60t)^2 + (-48t)^2 under the square root.

Then you should square both sides and simplify and get

5904t^2 = 529

But this gives the time in hours. The question asks for the minutes, so convert the number you get here into minutes...

 

[CathyLou]

Thanks so much for your help.

I got 18 minutes.

Cathy

 

[learningphysics]

Thanks so much for your help.

I got 18 minutes.

Cathy

Yup, that's it. no prob.