Average velocity (magnitude) in meters per second?

Question

I go out for a walk. I first walk 1km north, then I walk 2 km west. Then I walk northwest ( an angle exactly between north and west) for 2 km. What is my displacement (magnitude and direction (direction means find the angle))? If I did this walk in 2 hours, what is my average velocity (magnitude) in meters per second?

Answer 1

This is very easy to do if you plot the movements graphically on paper. Use 10cm on paper as 1km. Draw it out and then you measure from the starting point to the ending point to get the total displacement and use a protractor to measure the angle.

If you want it to be more precise, you have to break the movements into north and west components. You would then have a right triangle from the origin and you could figure out everything pretty easily.

The first two legs of the trip are easy. For the final leg, by walking 2km NW, you could also say that you basically walked sqrt2 km W and sqrt2 km N. The total N component of the walk is 1+sqrt2 and the total W component is 2+sqrt2. The displacement of that right triangle is the hypotenuse of that right triangle:
sqrt[(1+sqrt2)^2 + (2+sqrt2)^2], which is approximately 4.18km That makes an angle of arctan [(2+sqrt2)/(1+sqrt2)]. That's about 54.7deg north of west.

The average velocity is basically the total distance you actually walked divided by the time, 5km/2hr.

Answer 2

First you'll be able to desire to calculate the acceleration of the cart. it is in many cases finished with a hardship-loose kinematics equation, v = v_0 + at v = velocity v_0 = preliminary velocity a = acceleration t = time (Your very final velocity is 0, so your acceleration might desire to be unfavorable. besides the undeniable fact that, you in basic terms might desire to handle its absolute fee in this concern.) purely rearrange for acceleration. then you definately can calculate the tension with Newton's 2d regulation: F = ma F = tension m = mass