Vector Question.....

[arnab]

Any help with this question?

 

At time t = 0, a football player kicks a ball from the point A with position vector (2i + j) m

on a horizontal football field. The motion of the ball is modelled as that of a particle

moving horizontally with constant velocity (5i + 8j) m s–1. Find

(a) the speed of the ball,

 

 

 

(b) the position vector of the ball after t seconds.

 

The point B on the field has position vector (10i + 7j) m.

 

© Find the time when the ball is due north of B.

 

At time t = 0, another player starts running due north from B and moves with constant

speed v m s–1. Given that he intercepts the ball,

 

d) find the value of v.

 

I worked out part a, b, c but have no idea as to how to do part d :'(

[Greg H.]

My suggestion would be to keep in mind that if both objects (the ball and the player) start moving at the same time (t=0) and arrive at the same place in the same amount of time (which you found in answer c), then the velocity they travel is dependent on the distance they need to cover to reach that point.

[imatfaal]

There may well be a more elegant solution but I would initially work out the postion of point C - being the point at player running north intercepts the ball. You can then work out the distance moved and the time taken - and as it is a constant velocity that's all you need to know.

[Daedalus]

The following graph allows us to visualize the problem:

 

 

[math]\vec A[/math] is the position vector for where the ball was kicked, [math]\vec v_b[/math] is the velocity vector of the ball, [math]\vec B[/math] is the position vector where the player starts running north, [math]\vec v_p[/math] is that player's velocity vector, and [math]\vec C[/math] is the position vector where the player intercepts the ball.

 

You already know parts a - c:

 

a.) The speed of the ball is equal to the magnitude of [math]\vec v_b[/math] such that:

 

[math]|\vec v_b|=\sqrt{5^2+8^2}=\sqrt{89}\, m/s[/math].

 

b.) The position vector of the ball after [math]t[/math] seconds is:

 

[math]r_b(t)=\vec A + t \, \vec v_b=(2+5t)i+(1+8t)j[/math].

 

c.) The time when the ball is due north from [math]\vec B[/math] is when the [math]i[/math] component of the ball's position vector equals the [math]i[/math] component of [math]\vec B[/math]:

 

[math]2+5t = 10\, m, \ \ t=8/5 \, s = 1.6\, s[/math]

 

Now for part d:

 

1.) You already know it takes [math]1.6\, s[/math] for the ball to be located due north of [math]\vec B[/math].

2.) You also know the position that the player at [math]\vec B[/math] begins running for the ball.

3.) You also know that the player runs due north such that the position vector of the player after [math]t[/math] seconds is:

 

[math]r_p(t)=\vec B + t \, \vec v_p=10i+(7+v\, t)j[/math]

 

Now all you need to do is set the position vector of the ball [math]r_b(t)[/math] equal to the position vector of the player [math]r_p(t)[/math] at time [math]1.6\, s[/math] and solve for [math]v[/math]:

 

[math](2+5(1.6))i+(1+8(1.6))j = 10i+(7+v(1.6))[/math]

 

[math](2+5(1.6))i = 10i[/math]

[math](1+8(1.6))j = 13.8j = (7+v(1.6))j[/math]

 

The rest is up to you : )

Edited May 15, 2012 by Daedalus

[imatfaal]

Great post Daedalus - if anything a bit too much on the answers. This is homework help

[Daedalus]

Sorry about that. I based my reply on the assumption that he knew parts a - c. I'll trim it back next time : )

Edited May 16, 2012 by Daedalus