# I need sumone to help me on a few probs

1. Dec 14, 2004

### Kimorto

I have finals at school tomorrow for A.P. Physics (college level physics)
I was wondering if someone could help me for a few minutes and im just gonna ask about a few questions at a time.

(These questions are multiple choice, and I have all the answers from the teacher but I need help solving them, because I'm haveing a hard time where to start and which formulas to use.)

Question #1

The minimum takeoff speed for a certain airplane is 75 m/s. What minimum acceleration is required if the plane muse leave a runway of length 950 m? Assume the plane starts from rest at the end of the runway.

I thoughtt to use VF^2=vi+2ad but its not working any suggestions?

2. Dec 14, 2004

### learningphysics

3. Dec 14, 2004

### cyby

Vf^2 = vi^2 + 2ad. Solve for a. You had the wrong equation.

4. Dec 14, 2004

### dextercioby

Of course it's not working,coz it's wrong.
Use these ones:$d=\frac{at^{2}}{2}$ and $v_{f}=at$
Or the one u were trying to use written correctly (it can be deduced from the 2 i have posted).
Daniel.

5. Dec 14, 2004

### Kimorto

oops... I just noticed i plugged in something wrong, thanks for help on that problem. and the vi^2 i had that written down but I was reading off my work not the equation, and since it is at rest vi = 0. Alright... Next Problem.. These aren't numbered by the worksheet I'm just writing down the ones I don't know.

Problem #2
A cheetah is walking at a speed of 1.10m/s when it observes a gazelle 41.0m directly ahead. If the cheetah accelerates at 9.55 m/s^2, how long does it take the cheetah to reach the gazelle if the gazelle doesn't move?

I used D = Vi+(1/2)at^2 and manipulated it.
(2(d-Vi))/a = t^2
I came out with 2.89 Will someone check that for me, The answer is kind of blurred and the first number only has the top part on it. It looks like a 2.82s but I can only clearly see the part that says .82s.

6. Dec 14, 2004

### JinSu

Its x = xi + Vi(t) + (1/2)a(t)^2

7. Dec 14, 2004

### cyby

Again, wrong equation.

D = Vi*t+(1/2)at^2 should be the right equation. Always check your units to see if your equation makes sense!

8. Dec 14, 2004

### dextercioby

Again you're using wrong formulas.It should be
$$D=v_{i}t+\frac{at^{2}}{2}$$

So it's a second order algebraic equation which should have only one solution physically acceptable.

9. Dec 14, 2004

### JinSu

Yes, that dimension test with [L] [M] [T] works wonders for me.

[L] = [L] + [L/T][T] + C[L/T^2][T^2]

Last edited: Dec 14, 2004
10. Dec 14, 2004

### Kimorto

erg.. I hate this... No wonder i'm having a hard time in this class, I keep using the wrong equations. Thank you So much... These questions seem so easy.. I just haven't done any in a few months.

Alright now I have the right answer.

This is my last question for a about 20 minutes I have to do some more problems then I will ask more questions because I'm going to go over all the equations I have.

Question #3
A rock is thrown vertically upward from the surface of the earth. The rock rises to some maximum height and falls back toward the surface of the earth. Which one of the following statements concerning this situation is true if air resistance is neglected?
A)As the ball rises, its acceleration vector points upward.
B)The ball is a freely falling body for the duration of its flight.
C)The acceleration of the ball is zero when the ball is at its highest point.
D)The speed of the ball is negative while the ball falls back toward the earth.
E)The velocity and acceleration of the ball always point in the same direction.

The teacher said it is B... I don't understand why

I understand why D and E couldn't work... But why couldn't A and C work?

11. Dec 14, 2004

### JinSu

I know that A doesn't work because A is acceleration, which is gravity. If A was a positive vector, then your rock would zoom into the air and never come back.

g = -9.81 m/s^2

And I believe that the accelation for choice C is also gravity...

Not sure about the [C] one, I find this post interesting because I have an exam on it on Thursday (PHY for Engineers). If you don't mind posting the answer so I can maybe pick up some stuff for my exam just incase a possible trick question might appear.

12. Dec 14, 2004

### cyby

A is definitely not true. The moment the rock leaves your hand, it is in free fall - the only force acting on your rock is gravity. Since net force is downwards, so is acceleation.

C is definitely not true either. It is true that while on top of its path, the velocity is zero. However, gravity never stops acting on it!

Because gravity is the only force acting on the body throughout its path, that satisfies the definition of a freely falling body. Therefore, the answer is B.