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If your approach is right, then why this approach is not right as shown in the image :

scientia wrote:

Suppose at timetthe dog is running at angleθto the cat's direction of motion. The component of the cat's velocity parallel to that of the dog isucosθso the velocity of the dog relative to that of the cat iswhere

.) Suppose the dog catches the cat after timesthe separation between the animals at timet. (Note that the RHS is positive asT. Integrating givesNow, the component of the dog's velocity parallel to that of the cat is

vcosθ, so when the dog catches the cat we haveSubstituting the integral into the previous equation gives

This solution is copied from physicsforum website, and is non-calculus based, as integration is cancelled out.

Thanks for the reply. Hopefully I will make some big leap forward with this thing.

Please see my approach so far. I have used very small interval dt when cat has moved a distance of udt and dog vdt.

I am stuck with strange equation.

Thanks in advance

**pokrate**- Replies: 3

I always wondered about dx.dx ? How to handle that ? If I consider a small triangle with height = dy, and width dx, then

hypotenuse dh = sqrt(dy^2 + dx^2). Now suppose I want to differentiate this equation wrt time t, what we will get ?

**pokrate**- Replies: 10

A cat sitting in a field suddenly sees a standing dog. To save its life, the cat runs away in a straight line with speed u. Without any delay, the dog starts with running with constant speed v>u to catch the cat. Initially, v is perpendicular to u and L is the initial separation between the two. If the dog always changes its direction so that it is always heading directly at the cat, find the time the dog takes to catch the cat in terms of v, u and L.

**pokrate**- Replies: 1

Today while solving a boat / river problem, I wasn't convinced of vector drawing to find velocities. I was thinking of force acting on the boat which affects its position and velocity. But I couldn't think clearly of FORCE equation to find displacement, accelerations etc. Please help me. Assume river velocity : VR, Boat still water velocity : VB, Distance from Point A to B = D (straight distance from bank to bank). Now please explain how to find position of the boat in the form r = f(Ө) .

anonimnystefy wrote:

Hi pokrate

The acceleration of the object from the problem is the consequence of the gravitational acceleration. The gravitational acceleration can be split into two accelerations, one perpendicular to the inclined plane and one parallel to it, both less than g itself. The perpendicular part doesn't have any effect on the object, while the parallel part is the acceleration you want. The two are represented in Bob's picture, from which we see that a=g*sin(theta).

I like the explanation, as if something can be split into a direction, then it must be able to split it into other direction which is not possible in my case. So, that is the reason why a = g.sin(theta). Thanks for the simple explanation.

Now to wet some appetite :

Attached is a figure showing grooved inclined plane, of which side view will be like the figure we are discussing and also attached in my first post. AC = 4, BC = 3, and the angle of the grooved path is @, now if the ball is released from rest from the top of the plane, what will be its velocity at point B.

bob bundy wrote:

My approach is to draw a diagram first. Sometimes I find lines get in the way so it may take a few re-draws before I'm happy.

Then I put on all the forces (rather than accelerations) because when you have friction as well you'll need F and the normal reaction, R, and these are forces. If I'm helping someone on-line, I stick to {diagram in black or dark blue}, and {forces in red} so they stand out.

Then I resolve parallel and perpendicular to the slope to get the equations.

I also write down what I else know, like u = 0, s (for distance) or t (for time) = whatever, acceleration = gsin(a) and what I want to know; in your case v.

Then I can choose the right formula; v = u + at or v^2 = u^2 + 2as, depending on the rest of the question.

If you get stuck on one, you are welcome to post it here and I'll try to give some hints.

Bob

I guess you haven't taken a look at the figure I posted. Using that diagram I just want to see how a = g.sin(theta) ? Because using the frame I posted it is easy to see that a = g / sin(theta) and not g.sin(theta).

bob bundy wrote:

hi pokrate,

Welcome to the forum.

I've made you a fresh diagram (see below).

I've put the acceleration due to gravity in red.

My diagram software won't let me do theta so I've used 'a' for the angle of the slope.

The dotted lines show directions that are parallel and perpendicular to the slope.

You then have to split

ginto two components; one parallel to the slope and one perpendicular to the slope.Notice that the angle a is repeated in the red 'acceleration' triangle.

So the component you want is

g sin(a).Hope that helps.

Bob

ps. I've never found I needed coordinates at all for problems like these. Only time I've used them is to show that an object projected at an angle, and flying through the air, follows a parabolic path (if you ignore wind resistence).

If I read my textbooks first and then try to solve such problems, I can solve like you. But if I try to take a look at this problem from a fresh perspective, then I am still looking for a meaningful answer.

**pokrate**- Replies: 7

Hi,

This is my first post, so bear with me please.

I was solving normal friction less inclined plane problem and was trying to find velocity at the bottom of the plane.

All texts focus on using co-ordinate system with X axis along the plane, and 0,0 at the position of the block lying at the top, and Y axis perpendicular to the inclined plane.

I was trying to use 0,0 as the bottom of the inclined plane, and I get a = g/sin(theta) which means a = infinite for theta = 0.

Please help me !!!

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