# A kinematics problem in which direction of velocity keeps changing.

1. Oct 1, 2014

### Satvik Pandey

1. The problem statement, all variables and given/known data
A dog sees a rabbit running in a straight line across an open field and gives chase. In a rectangular coordinate system (L,0), assume:

(i) The rabbit is at the origin and the dog is at the point at the instant the dog first sees the rabbit.

(ii) The rabbit runs up the y-axis and the dog always runs straight for the rabbit.

(iii) The dog runs at twice the speed of the rabbit.

Assuming L=300 units, how much does the dog have to travel to catch the rabbit?

Enter your answer as the distance the dog has traveled from sight to meal.

2. Relevant equations

3. The attempt at a solution

From the figure
in vertical direction
$-dx=(2v-vcos\theta )dt$
$\int _{ 300 }^{ 0 }{ -dx } =\int _{ 0 }^{ t }{ (2v-vcos\theta )dt }$
$300=2vt-v\int _{ 0 }^{ t }{ (cos\theta )dt }$......................(1)
in horizontal direction
$vt=2v\int _{ 0 }^{ t }{ (cos\theta )dt }$
$\int _{ 0 }^{ t }{ (cos\theta )dt } =\frac { t }{ 2 }$
putting this value in eq (1)
$300=2vt-\frac { vt }{ 2 }$
$t=\frac { 200 }{ v }$
I think we have to find displacement of dog.
Distance traveled by rabbit in this time is 200
By pytha
Displacement should be $\sqrt { 40000+90000 } =360.55$
But this is incorrect.
Where did I go wrong.

2. Oct 1, 2014

### Orodruin

Staff Emeritus
How have you defined your dx? Just looking at your figure, the change in x should depend on $\sin\theta$ rather than $\cos\theta$.

3. Oct 1, 2014

### Satvik Pandey

$dx/dt$ is the rate at which separation between dog and rabbit decreases.
Does this separation not decrease with $2v-vcos\theta$ ?

4. Oct 1, 2014

### Orodruin

Staff Emeritus
This is true. I was confused by your use of x as the variable when you also had an x axis and I do not see something else strange in that approach at first glance.

However, in the last step, you seem to be trying to apply Pythagoras' theorem, but the path travelled by the dog does not constitute the hypothenuse of a right triangle. Instead, ask yourself this question: If the rabbit has travelled a distance of 200, how far has the dog travelled?

5. Oct 1, 2014

### BvU

It does. And initially it is 300. But the integral along the path of the dog is not 300. Only the vertical component is 300. In other words: initially dx is what you say it is, but as soon as theta is non-zero, the vertical component changes at a different rate!

6. Oct 1, 2014

### Satvik Pandey

As dog velocity is twice the velocity of rabbit so it should cover $2(200)$.
.

7. Oct 1, 2014

### Orodruin

Staff Emeritus
What he has done is to integrate the total gain of the dog on the rabbit, i.e., he is integrating to find the total distance gained, not the total distance travelled (x the distance to the rabbit, not the x coordinate). He is also computing the total horizontal distance travelled which then gives him the integral he was missing for the gain.

So is 2x200 = 400 more in line with the expected answer?

8. Oct 1, 2014

### BvU

Savik has done a very good job up to the point t = 200/v.

9. Oct 1, 2014

### Satvik Pandey

Thank you BvU and Orodruin.:)
I thought that the displacement of dog is being asked that's why I used Pythagoras theorem.I should have read the question more carefully.

Last edited: Oct 1, 2014
10. Oct 2, 2014

### ehild

Anyway, your method was ingenious, Satvik!

ehild

11. Oct 2, 2014

### Satvik Pandey

Thanks ehild. Receiving a like from you means a lot to me.:w
I have seen similar types of question which involves this approach to the solution in some book earlier.I just applied this method here.

12. Oct 2, 2014

### Orodruin

Staff Emeritus
I agree. +1

13. Oct 2, 2014

### Satvik Pandey

Thanks Orodruin. Receiving a like from you also means a lot to me.:w