# Homework Help: Motion Questions

1. Mar 12, 2010

### Paymemoney

Hi
I have a few question i don't understand how to complete.

1. The problem statement, all variables and given/known data
A model rocket leaves the ground, heading straight up at 49m/s.
a)What is the maximum altitude?What are its speed and altitude at b)1s c)4s

2. Relevant equations ( For all questions)
Constant acceleration formulas
Average speed equation

3. The attempt at a solution
Question a) i understand how to complete, however how would i solve b)

This is what i have tried:

t=1s
v-final=?
v-initial=0
a=9.80

used the v-final=v-initial + at
v-final = 9.8m/s

this is incorrect what have i done wrong?

1. The problem statement, all variables and given/known data
A jetliner leaves San Francisco for New York, 4600km away. With a strong tailwind, its speed is 110km/h. At the same time, a second jet leaves New York for San Francisco. Flying into the wind, it makes only 700km/h. When and where do the two planes pass each other?

3. The attempt at a solution
Well firstly i tried to find the time from NY to SF and NY to SF.

So i get 4.18hrs and 6.57hrs. I minus the figures to get the time that the two planes pass which is 2.37hrs. However according to the book's answer this is incorrect.

For the distance i went 400(1100-700) * 2.39 = 956km.

1. The problem statement, all variables and given/known data
A hockey puck moving at 32m/s slams through a wall of snow 35cm thick. It emerges moving at 18m/s. a) How much time does it spend in the snow? b) How thick a wall of snow would be needed to stop the puck entirely.

3. The attempt at a solution

i found question a) was t=0.014s, but could not find the correct answer to question b)

I have tried doing the following:

$$x= \frac{v-initial - v-final}{2} * t$$

$$x=\frac{32}{2} * 0.014$$

x=0.224m

P.S

2. Mar 12, 2010

### thebigstar25

in the question it is given that v-initial = 49m/s , so why in part(b) you said it is zero??
another thing, when you are considering such motion , you will need the following equations:

v-final = v-initial - gt , (notice the minus sign befor g, since the gravity always downward)

v-final^2 = v-initial^2 - 2gy (y is the height)

y = v-inital*t - 0.5gt^2

try again and tell us what you get ..

3. Mar 13, 2010

### Paymemoney

ok, i got the answers to the altitude but i cannot seem to get the right answer for speed.

For the first one with 1s i got 9.8m/s, answers says its incorrect.

4. Mar 13, 2010

### thebigstar25

are you sure that you substitue in with the right given value!!
please show me what equation you used , and what values you substitute there?..

5. Mar 13, 2010

### Paymemoney

i used

v-final = v-initial + at

where v-final is what i am trying to find.

so it would be: 49-9.8 which is 39.2m/s

before i made v-initial 0m/s which it fact it is 49m/s.

6. Mar 13, 2010

### thebigstar25

exactly :) .. maybe you didnt notice that i mentioned in my first post that in the question v-initial is 49m/s ..

7. Mar 13, 2010

### ideasrule

t isn't 0.014 anymore because the puck has to spend a longer time in the snow. Try calculating the acceleration of the puck, then using that to find the stopping distance.

8. Mar 13, 2010

### Paymemoney

so what would the time be, i got the v-initial=32m/s, v-final=0m/s and i need at least three values to find acceleration.

9. Mar 13, 2010

### rl.bhat

Using the equation
vf^2 = vi^2 + 2*a*s, find the acceleration of the puck in the snow. Then you can find the time which the puck takes to cross the ice.
Using this value of acceleration, and taking the final velocity as zero, you can find the thickness of the ice to stop the puck.

10. Mar 13, 2010

### Paymemoney

yeh, but to find this i need at least three values. i only got 2 how can i find the acceleration?:

v-final=0m/s
v-initial=32m/s
a=?
t=?
x=?

I can't define x(s) as some given value, because that is what i am eventually going to find.

11. Mar 13, 2010

### thebigstar25

for question(3) part(a) .. you are given v-initial 32m/s , v-final 18m/s and x = 0.35m to find the time:

1) use this equation (v-final^2=v-initial^2 + 2ax) to find a ..
2) then use (v-final = v-initial + at ) to find the time ..

for part(b) .. you are given v-initial 32m/s , v-final 0m/s , and you are asked for x=?

1) since you got the acceleration a from part(a) use it along with the equation (v-final^2=v-initial^2 + 2ax) to find x ..

hopefully this is clear enough :) .. try now once more ..

12. Mar 13, 2010

### Paymemoney

yep, that makes more sense and i have got the answer to be 0.512m

13. Mar 13, 2010

### thebigstar25

I believe you got the right answer now :) since obviously x in part(b) should be greater than 0.35 ..

14. Mar 15, 2010

### Paymemoney

so can anyone help me on my second question about the jetliner?

15. Mar 15, 2010

### rl.bhat

If they meet after time t at a distance x from New york, then
First jet travels a distance x km and the other jet travels (4600 - x). Velocities of jet is known. Find t and equate them. Then solve for x.

16. Mar 15, 2010

### Paymemoney

so would i use the v=d/t formula or the constant acceleration formula.
Because when i used the v=d/t formula i didn't get the correct answer.

17. Mar 15, 2010

### rl.bhat

The jets are not accelerating.
You have to write
t = x/v1 = (4600 - x)/v2.
Solve the equation to find x.

18. Mar 15, 2010

### Paymemoney

how can i find 't' if i only have the values of velocity?

If i used the total distance of 4600/110(v1), would this be correct?

19. Mar 15, 2010

### rl.bhat

You know the values of v1 and v2. Substitute them in
x/v1 = (4600 - x)/v2, and solve for x.

20. Mar 16, 2010

### Paymemoney

ok, this is what i have done:

305.56(4600-x) = 194.4(x)
1405576-305.56x = 194.4(x)
499.96x = 1405576
x = 2811.38m

book's answers is 1800km what have i done wrong?