How Do Velocity and Motion Principles Apply in Complex Physics Problems?

[spyroarcher]

Homework Statement

It's quite a handful but:
1. A train is moving east at 30m/s along a long, straight section of track. A person on the train has a dart gun that can shoot rubber-tipped darts at a speed of 10m/2. If the person aims the gun at an angle of 30 degrees north of east (I think it means 30 degrees above the horizontal?) and pulls the trigger, what will be the magnitude and direction of the velocity of the dart as it leaves the gun, relative to a person on the ground watching the train pass by?

2. A projectile is fired from a balcony 20m high at an angle of 30 degrees above the horizontal. It strikes the ground 10.4sec later.
a)What is initial speed.
b)What is magnitude and direction of the velocity vector after 7 seconds.
c) With the balcony as the origin, find the x and y coordinates of the highest point of the motion.

3.An object moves in two dimensions with a position vector of:
vector r(t)=(4+3t^2-(1/3)t^3)i+(t^2-5)j
a) Find magnitude and direction of the velocity at t=1s.
b)Find the position of the object when the x-component of the velocity is maximized.
c)Find the angle between the velocity vector and acceleration vector at t=1s.

4. A ball on a string moves in a vertical circle of radius R at a constant speed with a centripetal acceleration a. At its lowest point, the ball is negligible height above the ground. After several revolutions, the strong breaks at the highest point in the motion. In terms of R, a, and g, find the horizontal displacement of the ball from the time the string breaks.

Homework Equations

v(x)=V(i)cos(theta)
V(y)=V(i)sin(theta)
(The only ones i used so far)

The Attempt at a Solution

I believe I got the first 2 questions right, but I'd like to make sure that I did, then the rest kind of got me confused.
1. 40m/s, 30 degrees (I'm paranoid about this answer and does direction=theta?)
2a.98.07m/s
2b.87.158m/s, 12.97 degrees
2c.(5, 122.69)
3a. sqrt(29) and 21.8 degrees
3b. This is where I kind of got confused, so I derive the vector to (6t-t^2)i+(2t)j, and found that t=3. So do I then just plug 3 in for the position vector and get my answer?
3c. 52.76 degrees (I most likely did this one wrong)
4. 0? I mean i think its saying that the string breaks off when it's at its highest point, making it shoot a straight 90, therefore no horizontal displacement, but I am very sure that I am reading this question 100% wrong.
Thanks all for helping me.

 

[Mindscrape]

Without actually having done the problem, I'll give you what I can.

1 can't be right because the gun wasn't fired all in the same direction, the total velocity has to be less than the sum of the two

2 don't know, maybe okay.

Your work for 3b is okay, so you should get the right answer for that. 3a looks okay. No idea about 3c.

For 4, it's actually the opposite. If the string breaks off at the highest point all the tangential velocity is pointing in the horizontal direction.

 

[spyroarcher]

Wait is the x component going to be 40? and how is it not 30 degrees

 

[Mindscrape]

You have to use galilean relativity on both x components to find the total x component. Then you you will have the correct y component, and then you can find the total velocity. Looks to me, with mental math, like the answer should be ≈39 and the angle would be arctan(5/~38.6) (not 30º, a bit weird huh? gets weirder in special relativity).