1. Sep 12, 2004

### Shay10825

Hi everyone! I really need help on these problems! I'm so confused . Sorry this post is so long and they are numbered weird.

6.) A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s^2. At an altitude of .5 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

My work:

Vi= 0, a=10 m/s^2, d(or x) = 500 m

Now I'm stuck! I know you have to split it into 2 parts (one on the way up and one on the way down), and on the way down a=-9.8. But you don't know Vf or t so I don't know what to do.

10.) A boy on a skate board skates off a horizontal bench at a velocity of 10 m/s. One tenth of a second after he leaves the bench to two significant figures the magnitude of his velocity and his acceleration are what?

My work:

Would you put the information into vectors? and if so how? Did they give enough information?

ANSWER: 10 m/s ; 9.8 m/s^2

13.) A juggler throws two balls to the same height so that one is at the halfway point going up when the other is at the halfway point coming down. At that point:

a. Their velocities and accelerations are equal.
b. Their velocities are equal but their accelerations are equal and opposite.
c. Their accelerations are equal but their velocities are equal and opposite.
d. Their velocities and accelerations are both equal and opposite.
e. Their velocities are equal to their accelerations.

How do you know that the answer is c?

14.) A particle starts from the origin at t=0 with a velocity of 8j m/s and moves in the xy plane with a constant acceleration of (4i + 2j) m/s^2. At the instant the x coordinate of the particle is 29 m, what is the value of its y coordinate?

My Work:

xi= 0 , Vix=0, ax=4, xf=29, Viy=8, ay=2, yf=?, Vfy=0

0= 64+4d
-64=4d
d=-16
yf=-16

But this is not the correct answer :surprised :grumpy: ! It's not even close to any of the choices . ahhhhhhh!!

15.) A particle starts from the origin at t=0 with a velocity of 6i m/s and moves in the xy plane with a constant acceleration of (-2i + 4j) m/s^2. At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin?

???????????????????
I don't know where to start. I don't understand how you would find this.

16.) A ball is thrown horizontally from the top of a building .1 km high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?

17.) A rifle is aimed horizontally at the center of a large target 60 m away. The initial speed of the bullet is 240 m/s. What is the distance from the center of the target to the point where the bullet strikes the target?

You have 3 vectors that form a triangle and you need to find the third side. 60 is one side and 240 is the other. I found the angle between them to be 75.52. Then I used the law of cosines and found the third side to be 232.379. This is not even close to any of the choices .

19.) A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the acceleration of a passenger at his lowest point during the ride?

My Work:

ar= (V^2)/(r)

C= 30pi so v= [(5*30pi)/1 min]
V = 150 m / 60 s
so: ar=.027

21.) Two cars are traveling around identical circular racetracks. Car A travels a constant speed of 20 m/s. Car B starts at rest and speeds up with constant tangential acceleration until its speed is 40 m/s. When car B has the same (tangential) velocity as car A, it is always true that:

a. it is passing car A
b. it has the same linear (tangential) acceleration as car A.
c. it has the same centripetal acceleration as car A.
d. it has the same total acceleration as car A.
e. it has traveled farther than car A since starting.

Any help would be greatly appreciated
~Thanks

Last edited: Sep 12, 2004
2. Sep 12, 2004

### Parth Dave

6) Well one thing is that Vi is not equal to zero. When x = 500 m, Vi is not 0. You're right, you can break this up into two parts. The first patr would be what the rocket does in the first 500 m. The second part would be what the rocket does until it gets to its maximum height (remember, you dont care what happens to the rocket on the way down, you are only concerned with how high it gets). Well if we look at part two, xi (initial displacement) = 500 m, however, a is not 10 ms^-2. It is actually -9.8 ms^-2. This is because the engine cuts off and the rocket doesn't generate any upward thrust. Also, when the rocket reaches its highest point, the velocity will be 0 (ie Vf = 0). When the rocket's velocity reaches 0, it is at its highest point. After that the velocity will become negative and the rocket will go down.

So for part 2, Vf = 0, a = -9.8 ms^-2, xi = 500 m, Vi = ?.

The only thing you don't know is the initial velocity. You can find the initial velocity from in the first part. Remember, the final velocity of the first part will be the initial velocity of the second part.

3. Sep 12, 2004

### Parth Dave

10. This is similar to a car driving off a cliff. You know what his velocity and acceleration (which happens to be 0) in the x direction is and the acceleration (-9.8 ms^-2) and initial velocity (0 ms^-1)in the y direction. So just find the velocity in the x and y direction after 0.1 s and than find the magnitude. Repeat for acceleration.

4. Sep 12, 2004

### Leong

Question #10

$$\vec{a}=\frac{\vec{v}-\vec{u}}{t}$$
$$\vec{u}=10\vec{i}$$
$$-9.81\vec{j}=\frac{\vec{v}-10\vec{i}}{\frac{1}{10}}$$
$$\vec{v}=10\vec{i}-0.981\vec{j}$$
$$|\vec{v}|=10\ m/s$$
$$\vec{g}=-9.81\vec{j}$$
$$|\vec{g}|=9.8 \ m/s^2$$

5. Sep 12, 2004

### Leong

Question #19

v=r*w
$$w=\frac{5*2\pi}{60}\ rev/s$$

Last edited: Sep 12, 2004
6. Sep 12, 2004

### Leong

Question #14

You can't assume $$v_{yfinal}=0$$
Use
$$s_{x/y}=u_{x/y}t+\frac{1}{2}a_{x/y}t^2$$

Find t for $$s_x=29$$

Substitute the time t to find $$s_y$$

7. Sep 13, 2004

### Shay10825

What does w r and v represent?

Last edited: Sep 13, 2004
8. Sep 13, 2004

### amwbonfire

Question 19

v=Velocity, in m/s
w=time (I use 't', i've never seen 'w' used before...)

I think this solution to this problem was completed incorrectly. Here's how it should be done:

Use
$$v=circumference/t$$
where t=time

Remember that circumference = 2 x pi x r

r=2.pi.r
t=60/5=12
v=?

Find v using the formula.

Now use this formula to find the acceleration:

$$a=v/r$$
where v=average velocity.

You're not finished yet though. You'll need to add this acceleration to gravity (9.8ms) because gravity also acts downwards. So you'll get a + 9.8

Andy
AMW Bonfire

Last edited by a moderator: Sep 13, 2004
9. Sep 13, 2004

### amwbonfire

Any others you need help with?

Andy
AMW Bonfire

10. Sep 13, 2004

### amwbonfire

Don't worry too much about the numbers in this question, they're just there to throw you off. The question is say, when both cars are travelling at the same speed, 20m/s, which of (a,b,c,d,e) is true?

Use the process of elimination to solve multi choice questions.

a) is wrong because B cannot pass A if they are both travelling at the same speed. Hence this answer is wrong.

b) The questions states that car A has no linear acceleration, while car B does. Therefor, they can't be equal. Hence this answer is wrong.

c) Centripital acceleration acts towards the center of the circle, and is given by the formula, a=v/t. If the velocities are the same, then the accelerations must be the same. Hence, this answer is correct.

d) Total acceleration means centripital acceleration + linear acceleration. Both centripital accelerations are equal (same velocity, see part c). However,
their linear (tangential) accelerations are different (see part b). Hence this answer is wrong.

e) This is making an assumption, which you can't do. Hence this answer is wrong.

So, c) it is!

I must admit, this question was pretty difficult. It could have been worded a lot better.

Andy
AMW Bonfire

11. Sep 13, 2004

### Shay10825

Thanks
yeah i need help with 13, 15, and 17

12. Sep 13, 2004

### amwbonfire

Ok, I'll do them now. :tongue2:

So are you learning all this stuff now, in physics? I'm only a bit ahead of you, I learnt this stuff a few months ago.

13. Sep 13, 2004

### amwbonfire

13

a) Wrong - their accelerations are equal. Gravity is accelerating the one going downwards, and deccelerating the one going up, and gravity is a constant. Since gravity is the only accelerating force acting on them, their acceleration is equal, (9.8m/s^2 downwards.)

However, their velocities are not equal. They are opposite (one is going up, one's going down.)

b) Wrong - See a) Velocities aren't equal, they're equal and opposite

c) Correct - See part a) and b)

d) Wrong - See the above parts

e) Wrong - See the above parts

Make sense?

Andy
AMW Bonfire

14. Sep 13, 2004

### amwbonfire

15

Don't panic!

Whenever you get a calculation, always write down what you know. It makes everything easier to see.

v(initial)=6i m/s
v(final)=?
a= (-2i + 4j) m/s^2
starts at coordinate (0,0)
need to find max x point

Sorry, I've run out of time. I'll help you some more, when I get home. But try and see where you can go from here. You need to find where x is greatest. Instead of x, think of it as displacement, then use the equations of motion to work out where displacement is greatest. Then you're done!

Andy
AMW Bonfire

15. Sep 13, 2004

### amwbonfire

17

All I can suggest is review your trig.

I'll have a look when I get home.

Andy
AMW Bonfire

16. Sep 14, 2004

### Leong

Question #19

I guess i have to find alternative since Latex has not been working since yesterday. Have a look at the attached file. Forget about the 'w', i don't want to confuse you. just use the circumference given by 2*pi*r.

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17. Sep 14, 2004

### Leong

Question #17 : 2D Projectile Motion

You can't use a right triangle for a projectile motion.

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18. Sep 14, 2004

### Leong

Question #15

Have a look at the attached file.

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19. Sep 15, 2004

### doyle_43

Hi People.

As you can tell I’m new here, and I’m having problems with my Physics homework.
I have missed a few of my lessons recently due to illness, and now it’s all mumbo jumbo.

Anyway, I kind of understand this, but I’m still getting a few wrong.

(Sorry that I have posted in this thread, but its stupid posting the same topic in a different thread.)

Having problems with this:

(6) A rocket is fired vertically upwards with an initial velocity of 100m/s^2. Calculate:

a. the time for the rocket to reach its greatest hight >answer> 10 seconds
b. the hight to which the rocket rises >answer> 500 meters
c. its velocity after 4.0 seconds >answer> 60 m/s^2
d. the acceleration at the top of the path of the rocket >answer> 9.8 m/s^2
e. when the rocket is 400 meters above the ground. >answer> 5.5 seconds

If anyone could explain how to get these answers, it would be much appreciated. :uhh:

I am getting very confused with the Kinematics Equations.

Thanks again for anyone who can help!

-- doyle_43

20. Sep 15, 2004

### Leong

Doyle,
Have a look ....

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