Physics, Vectors and resultant vectors

Question

1) A dog is on a street chasing cars. He runs 30m west and stops. He then chases another car 20m north before chasing the mailman 60m East.

..a) What is the dogs distance traveled?

..b) What is the dogs resultant displacement?

Use a scaled vector diagram to find your answer (ruler and protractor)

And also

2)

Using the component method, find the magnitude of the resultant displacement fr a trip of 25kn East and 15500m south.

Can anyone pls help assist me with them? Not to sure of what to do in them

What will be the resultant displacement in question 1?

Answer 1

1) Drawing a graph is extremely useful for visualizing this problem and coming up with the correct answer. If we define the point (0,0) as the dog's starting point, we can plot points for each leg of the trip. This will help visualize the dog's travel path. We'll assume a standard convention of North = +y, East = +x, South = -y, West = -x.

Starting Point = (0,0)

30m West = (-30,0)

20m North = (-30,20)

60m East = (30,20)

a) Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. The dog went 30m West, then 20m North, then 60m East. So its total distance is 30m + 20m + 60m = 110m.

b) Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position. So we are only interested in the dog's final and initial positions, more specifically the change in this position. An easy example would be if the dog went 30m West, then 30m back East. It's total distance traveled would be 60m but its displacement would be 0m. We are right back where we started.

So if our initial position is (0,0) and final position is (30,20), how do we calculate the change in position? We draw a straight line between the points and calculate the length of this line. We can draw a right triangle and use Pythagorean's Theorem.

Disp = sqrt[(30-0)² + (20-0)²]

Disp = sqrt(900 + 400) = sqrt(1300)

Disp = 36.06m

The actual resultant displacement vector would be a line from (0,0) to (30,20), with an arrow at the (30,20) end. It's magnitude is 36.06m.

2) Similar to 1) part (b), we can plot a single point for our final position (25000, -15500) in meters or (25, -15.5) in kilometers. Use Phythagorean's theorem again to find the length of the line connecting (0,0) to (25, -15.5).

Disp = sqrt[(25-0)² + (-15.5-0)²]

Disp = sqrt(625 + 240.25) = sqrt(865.25)

Disp = 29.42km or 29420m