# Homework Help: Couple Kinematics/Vectors Questions

1. Aug 30, 2008

### ff4930

Hello everyone,

I have couple of online homework questions and I am not asking anyone to help me do the work. I will post the question and my answers to them and I would just like someone to verify if the answer is indeed correct because I only have one chance to submit my homework.

Determine the x and y components of the following vectors in the xy plane.
(a) A 14 m displacement vector that makes an angle of 38° clockwise from the +y direction.
x: 11.0m from 14*cos(38)
y: 8.6m from 14*sin(38)

(b) A 22 m/s velocity vector that makes an angle of 36° counterclockwise from the -x direction.
x: -17.8m/s from 22*cos(36) and negative because is in the negative quadrant
y: -12.9m/s from 22*sin(36) and negative because is in the negative quadrant

(c) A 35 lb. force vector that makes an angle of 171° counterclockwise from the -y direction.
x: 34.6lb.from 35*cos(171) and positive because is in the is 171 degrees
y: 5.5lb. from 35*cos(171) andpositive because is in the is 171 degrees

In doing a calculation, you end up with m/s in the numerator and m/s2 in the denominator. What are your final units?

1/s since m/m cancel and s/s2 = s

At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the roof. During their descent to the ground what happens to the distance between the two objects?

It is proportional to t?

A runner runs 2.0 km in a straight line in 9 min and then takes 25 min to walk back to the starting point. (The running and walking is in a straight line.)
(a) What is the runner's average velocity for the first 9 min?
2.0/9 km/h since the displacement is 2.0 and total time was 9 mins?

An object with an initial velocity of 7.0 m/s has a constant acceleration of 2.0 m/s2. When its speed is 17.0 m/s, how far has it traveled?

since is constant acc. each second it will travel 2.0m/s2
time 0 = 7.0 m/s
time 1 = 9.0 m/s
time 2 = 11.0 m/s
time 3 = 13.0 m/s
time 4 = 15.0 m/s
time 5 = 17.0 m/s

you add it up is 72 meters?

Thanks for your help and time.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 30, 2008

### tiny-tim

Hello ff4930!

Sorry

(b) is right but (a) and (c) are wrong, and so is 1/s.

(Going out now for the evening, so no time to look at the others)

3. Aug 30, 2008

### ff4930

4. Aug 30, 2008

### Astronuc

Staff Emeritus
ff4930, it would help if you write the details of the solution rather than just the answer.

In the first 3 problems a, b, and c on the Cartesian (x, y) system, pay attention to the reference axis (i.e. the axis from which the angle is taken). In problem a (+y), in b (-x) and in c (-y).

The answer given for (a): x: 11.0m from 14*cos(38), y: 8.6m from 14*sin(38) would be appropriate if the reference axis was +x (with +x from the origin and proceeding horizontally to the right - assuming the conventional righthanded system in which positive angles sweep counterclockwise).

In the case of numerator of (m/s) and denominator (m/s2), write it out.

For the two objects released simultaneously, write the equations for the altitude for both object y1(t) and y2(t) and take the difference. What is the difference?

5. Aug 30, 2008

### ff4930

For A is clockwise from +y so it should lay in the first quadrant no?
and x and y components should be positive.

Is the 14m displacement not the hypotenuse of the triangle trying to make?

Im not too familiar about the referencing y axis/xaxis part.

For M/S divided by M/S^2, you multiply with the numerator by the reciprocal of the denominator no? so It should be M/S * S^2/M which = to S?

For the altitude equation is it one of the equations that uses constant acceleration?

6. Aug 30, 2008

### Astronuc

Staff Emeritus
Yes, in the first quadrant, both would be +, but the angle in measured clockwise from the vertical, but when one uses x = r cos (theta) and y = r sin (theta), one is taking theta counterclock wise from the +x axis or horizontal. If one rotates the reference 90 degrees to +y, then sweeps the angle -theta, what becomes the relationship between x, y and theta (assuming r is fixed)?

Correct. But please don't guess. One needs to be sure.

Assume the same constant acceleration, but they start simultaneously at different elevations.

This might be a useful reference - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

7. Aug 30, 2008

### ff4930

If you rotate the reference 90 degrees to +y it would lay in 2nd quadrant where x component is negative and y component is positive? Im still alittle confused.

Since the acceleration is constant and they start simultaneously their accelerations are the same so their difference is still 10m apart?

I was wondering the previous 2 problem is correct or not.

runner runs 2.0 km in a straight line in 9 min and then takes 25 min to walk back to the starting point. (The running and walking is in a straight line.)
(a) What is the runner's average velocity for the first 9 min?
2.0/9 km/h since the displacement is 2.0 and total time was 9 mins?

An object with an initial velocity of 7.0 m/s has a constant acceleration of 2.0 m/s2. When its speed is 17.0 m/s, how far has it traveled?

since is constant acc. each second it will travel 2.0m/s2
time 0 = 7.0 m/s
time 1 = 9.0 m/s
time 2 = 11.0 m/s
time 3 = 13.0 m/s
time 4 = 15.0 m/s
time 5 = 17.0 m/s

you add it up is 72 meters?

8. Aug 30, 2008

### ff4930

Ok I managed to figured out some of the answers on my own.

Q-An object with an initial velocity of 7.0 m/s has a constant acceleration of 2.0 m/s2. When its speed is 17.0 m/s, how far has it traveled?

since is constant acc. each second it will travel 2.0m/s2
time 0 = 7.0 m/s
time 1 = 9.0 m/s
time 2 = 11.0 m/s
time 3 = 13.0 m/s
time 4 = 15.0 m/s
time 5 = 17.0 m/s
I got this one wrong it wasn't 72m, can someone tell me what I am doing wrong here?

The law of radioactive decay is N(t) = N0e-λt, where N0 is the number of radioactive nuclei at t = 0, N(t) is the number remaining at time t, and λ is a quantity known as the decay constant. What is the dimension of λ? (T = time)
a-1/T
b-T
c-T2
d-2/T
e-√T

I have no idea what this questions ask, can someone simplify it?

And I have no clue about the quadrant question regarding the x and y components.