Finding the Position of a Football in Motion: A Vector Analysis

[thereddevils]

Homework Statement

Footballer P kicks a ball to another footballer Q with an initial speed of 10 m/s in the direction of unit vector j.Because there is resistance of the field, the ball experiences a retardation of magnitude of 1 m/s^2. By assuming the position of player P as the origin and the movement of the ball as always horizontal and in a straight line

find the position of the ball at time , t.

Homework Equations

The Attempt at a Solution

a= -1

$$v=\int a dt=-t+c$$

when t=0 , v=10

therefore , v (resistance)= -t+10

so nett velocity of the ball = 10-(-t+10)=t

And in vector form , v=t^2 j

Am i correct ?

 

[CompuChip]

Homework Statement

Footballer P kicks a ball to another footballer Q with an initial speed of 10 m/s in the direction of unit vector j.Because there is resistance of the field, the ball experiences a retardation of magnitude of 1 m/s^2. By assuming the position of player P as the origin and the movement of the ball as always horizontal and in a straight line

find the position of the ball at time , t.

Homework Equations

The Attempt at a Solution

a= -1

$$v=\int a dt=-t+c$$

when t=0 , v=10

therefore , v (resistance)= -t+10

With that, I agree, but I don't follow you after that.
So the velocity of the ball is given by v(t) = -t + 10.
All you have to do is find the position.

 

[ehild]

1.

The attempt at a solution

a= -1

$$v=\int a dt=-t+c$$

when t=0 , v=10

therefore , v (resistance)= -t+10

Do you use v for velocity? What do you mean on v(resistance)?

so nett velocity of the ball = 10-(-t+10)=t

And in vector form , v=t^2 j

Am i correct ?

No. What do you mean on nett velocity? v(t)=t means that the velocity increases, but the problem said that the velocity decreased. Vector form includes the direction of velocity, and does not influence the time dependence.
One gets the velocity as function of time by integrating the acceleration, and taking initial velocity into account. You have done it, the velocity is v(t) = (10-t) j in vector form.
You get the position by integrating the velocity and taking the initial position into account.

ehild

 

[thereddevils]

thanks guys , i did mess it up .

Attempt no 2:


integrating the acceleration vector , v=10-t

so at time , t

position vector , r = (-t^2+10t) j

Better now ?

 

[ehild]

Better but still not correct. What is

$$\int{tdt}$$

?

ehild

 

[thereddevils]

Better but still not correct. What is

$$\int{tdt}$$

?

ehild

ok so integrate the velocity function once again , s=10t-t^2/2

and the position vector would be 10t-t^2/2 j

But why is my method wrong?

I thought of adding the original position vector of the ball to the velocity of the ball at time,t to get its final position vector . ie

s=(0i+0j)+(0i+(10-t)j)t=(10t-t^2 )j ?

 

[ehild]

I thought of adding the original position vector of the ball to the velocity of the ball at time,t to get its final position vector . ie

s=(0i+0j)+(0i+(10-t)j)t=(10t-t^2 )j ?

The position vector is the integral of the velocity vector, taking initial position into account. The displacement is velocity times time in case of uniform velocity, and it is not the case now.

ehild

 

[thereddevils]

The position vector is the integral of the velocity vector, taking initial position into account. The displacement is velocity times time in case of uniform velocity, and it is not the case now.

ehild

thanks !

 

[ehild]

Wait, x=10t-1/2 t^2 is not the end of the problem. The retardation force acts only till the ball moves. As soon as it stops, there is no velocity and no deceleration so the ball stays at the same place. The ball will stop in 10 s.
So the position is x(t)=10t-0.5t^2 if 0≤t≤10 s, and x(t)=50 m if t>10 s.

ehild

 

[thereddevils]

Wait, x=10t-1/2 t^2 is not the end of the problem. The retardation force acts only till the ball moves. As soon as it stops, there is no velocity and no deceleration so the ball stays at the same place. The ball will stop in 10 s.
So the position is x(t)=10t-0.5t^2 if 0≤t≤10 s, and x(t)=50 m if t>10 s.

ehild

ok ,this is the continuation of the problem.

(1) If player Q runs with velocity (-4i-3j) m/s , determine whether he can receive the pass from player P .

(2) If player Q runs in the direction -i with the same speed in (1) , determine whether he can receive the pass from player P.

So in this case , i think i will need to assume that the ball stops in 10 s, can I ?

(1) $$r_Q=(20-4t)i+(32-3t)j$$

If r_p=r_Q

$$(20-4t)i+(32-3t)j=(10t-0.5t^2)j$$

but after comparing ,the time for both equations don match , hence $$r_p\neq r_q$$

 

[ehild]

Where was Q initially? at point(20,32)?

ehild

 

[ehild]

Go ahead to the second question. It looks more promising :).

ehild

 

[thereddevils]

Where was Q initially? at point(20,32)?

ehild

yes i missed that part , so am i correct ?

For (2)

|Q|=5

r_q=(20-5t)i+32j

r_P=r_q

(10y-0.5t^2)j=(20-5t)i+32j

By comparing , 20-5t=0 , t=4

32=10t-0.5t^2

solving gives t=4 too.

Yes for this .

 

[ehild]

Well done!

ehild

 

[thereddevils]

Well done!

ehild

thank you ehild !