Guys, I have 4 easy questions. (probably easy to you)

[MoreZitiPlease]

Homework Statement

1. While descending at a constant speed of 1.0 m/s, a scuba diver releases a cork, which
accelerates upward at 3.0 m/s2. What is the diver’s depth when the cork reaches the surface 2.0 s later?

2. A car with a velocity of 27 m/s slows down at a rate of - 8.5 m/s2 to a stop in a distance
of 43 m on a dry road. The same car traveling at 27 m/s slows down at a rate of -6.5 m/s2 to a stop on a wet road.
a. How much farther does the car travel on the wet road before coming to a stop?
b. What maximum car speed will allow the car traveling on the wet road to stop in a distance of 43 m?

3. A car traveling at 14 m/s encounters a patch of ice and takes 5.0 s to stop.
a. What is the car’s acceleration?
b. How far does it travel before stopping?

4. An accelerating lab cart passes through two photo gate timers 3.0 m apart in 4.2 s. The velocity of the cart at the second timer is 1.2 m/s.
a. What is the cart’s velocity at the first gate?
b. What is the acceleration?

Homework Equations

WE NEED TO USE ONE OF THESE EQUATIONS FOR THESE QUESTIONS
V= Vo + at
d = Vo t + ½ at2
V2 = Vo2 +2ad
d = ½ (V + Vo) t

g= -9.8 m/s2

The Attempt at a Solution

I am sorry. I really have no idea. I have been sick for quite some time and missed a lot of school time. Since we are on break, this was our assignment, I cannot ask the teacher for help.

If you can tell me which equation I need to use, I am sure that will help.

 

[Retsam]

Well just use Newtons equations of motion, which are:

v = u+at

v^2 = u^2 + 2as

s = ut + 0.5a(t^2)

where u = initial speed, v = final speed, s = distance, t = time, a = acceleration

Then just need to look at each situation, break it down, and apply soem of the equations.

 

[MoreZitiPlease]

Well just use Newtons equations of motion, which are:

v = u+at

v^2 = u^2 + 2as

s = ut + 0.5a(t^2)

where u = initial speed, v = final speed, s = distance, t = time, a = acceleration

Then just need to look at each situation, break it down, and apply soem of the equations.

Thats my main problem. I don't know which equation to use.

 

[Feldoh]

Make a list of the variables you have and what you're trying to solve for, and find the equation that applies.

Take 3a) We know the time (t), the initial velocity (vo), and the final velocity (v). We're looking for acceleration. Which equation would you use?

 

[MoreZitiPlease]

For #1, should I be using
d = Vo t + ½ at2

??

 

[Feldoh]

For #1, should I be using
d = Vo t + ½ at2

??

To find the distance of the surface of the water to the diver when he releases the cork, yes.

 

[Retsam]

For example the first answer is as follows:

1. Find depth of the diver when cork is released

so we work out distance traveled by the cork with

a = 3
t = 2
v = ?
u = 0
s = ?

we want s so pick equation that has s as the only unknown in it

s = ut + 0.5a(t^2)

so

s = 0.5(3)((2)^2)
s = 6 m

Now find distance traveled by diver in the 2 seconds

s = ut as we know a=0
s = (1)(2) = 2m

So total depth is 6+2 = 8m

 

[MoreZitiPlease]

why would u be 0? Isnt the diver going at 1.0m/s

 

[MoreZitiPlease]

Alright, I solved it in my head but I just used my common sense tosolve the question. Simple mental math.

I did get 8. I just hate these formulas

idk about the other 3 tho

 

[Feldoh]

why would u be 0? Isnt the diver going at 1.0m/s

It's the starting velocity of the cork, not the diver. By solving the equation you're getting the distance to the diver at the depth where he releases the cork. You also need to take into account the distance he moves in the time it takes the cork to go up, that's where we get the extra 2m.

 

[MoreZitiPlease]

I figured it out. Now what about the next 3? I feelso ignorant now.

 

[Retsam]

Actually i made mistake :P

The cork does have an initla velcocity of 1m/s in the downwards direction as it is moving in the same direction as the diver. So first it must deccelerate before it can move in the surface direction. Taking upwards to be the positive direction we must use :

v = ?
u = -1 negative as it is in downwards direction
a = 3 positive as it is in upwards direction
t = 2
s = ?

s = ut + 0.5a(t^2)
s = (-1)(2) + 0.5(3)(2^2)
and so s = 4m

and so we get 2+4 = 6m depth

 

[MoreZitiPlease]

Actually i made mistake :P

The cork does have an initla velcocity of 1m/s in the downwards direction as it is moving in the same direction as the diver. So first it must deccelerate before it can move in the surface direction. Taking upwards to be the positive direction we must use :

v = ?
u = -1 negative as it is in downwards direction
a = 3 positive as it is in upwards direction
t = 2
s = ?

s = ut + 0.5a(t^2)
s = (-1)(2) + 0.5(3)(2^2)
and so s = 4m

and so we get 2+4 = 6m depth

Yea, but when you are descending into the water, you are going down. The diver would be 8m into the water.

edit-I am not sure myself now. You've got me thinking.

 

[Astronuc]

With respect to problem 1, it would probably be safe to assume the cork is release with zero initial velocity, since the deceleration in water would be very rapid (i.e. much faster than the 3 m/s². The diver could release it in such away to have a zero initial velocity.

The the problem is one in which the cork simply accelerates at a constant rate vertically (to the surface) for 2 sec, and the diver continues to descend at constant velocity.

 

[MoreZitiPlease]

so you're saying the answer is 8? What are your thoughts about the other problems

 

[Astronuc]

Use V² = V_o² + 2ad for problem 2.

One is given V_o and a. Find the distance in part a.

Then knowing the distance d and a, one can find V_o.

 

[MoreZitiPlease]

I am confused. I don't know which goes where

 

[Retsam]

Well V is your final speed which is going to be 0, then the Vo is the speed of the car before the braking occurs

 

[MoreZitiPlease]

For d on the dry road

I got 452m

I doubt that is right..

 

[Retsam]

Well you already know d for the dry road, what you need is d for the wet road, so :

V = 0
Vo = 27 m/s
a = -6.5 m/s2
t = ?
d = ?

so using V = Vo^2 + 2ad

-(Vo^2)/2a = d
-(27^2)/2(-6.5) = d = 56m

 

[MoreZitiPlease]

nvm

so the anserr is 13?

Now how wouild I solve part B?

 

[Retsam]

V = 0
Vo = ?
a = -6.5
t = ?
d = 43

V^2 = (Vo^2) + 2ad

V^2 - 2ad = Vo^2
(0)-(2)(-6.5)(43) = Vo^2

From there just squareroot to find Vo
Questions are kind of easy if you follow the same process for each one, noting down the variables you know and which you don't know, and then picking the relevant equation.

 

[MoreZitiPlease]

My answer was 23.64

for #3

I found the acceleration to be 2.8m/s

now how would I find its distance?

 

[MoreZitiPlease]

ok, I got 35m

Now, onto #4..

 

[Astronuc]

4. An accelerating lab cart passes through two photo gate timers 3.0 m apart in 4.2 s. The velocity of the cart at the second timer is 1.2 m/s.

V = Vo + at can be generalized to V₂ = V₁ + at,

and

V₂² = V₁² + 2ad


One is given the velocity at the second gate V₂ = 1.2 m/s, and the distance d = 3m and time 4.2 s, between the first and second gates. One may assume a constant acceleration unless told otherwise.

So one could use two equations above, and solve for V₁ and acceleration a.


Or one could solve for V₁ knowing V₂ and the average velocity between gates 1 and 2, which is just the distance divided by the time, and then use the first equation to find a.

Here is a good reference for this work - http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html

particularly - http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot1

I recommend reviewing the pages at Hyperphysics in order to better understand the physics.

 

[MoreZitiPlease]

im having trouble solving for the acceleration..

which equation did u use to find acceleration?

 

[Astronuc]

im having trouble solving for the acceleration..

which equation did u use to find acceleration?

Did you find V1.

If so, then the acceleration is simply (V2-V1)/t. One has to find V1 first.

 

[MoreZitiPlease]

I don't know how to find V1

 

[Astronuc]

To find the velocity at gate 1 in problem 4, one can generalize the equation

d = ½ (V + Vo) t => d = 1/2 (V2 + V1) t

or 2 d / t = V2 + V1 = 2 (d/t) - V2 = V1.

The value of d/t ( 3 m / 4.2 s) is simply the average speed over that distance. This is only valid for constant acceleration!

Then the acceleration is simply the change in velocity per unit time or

a = (V2 - V1)/t

 

[MoreZitiPlease]

iM SO LOST

MAYBE it is because I am reading these equations on a pC screen

cab y please just give me the answer? this problem has consumed 2 hours of my time

:(

 

[Astronuc]

See my previous post. The formulas and explanation are given.

V1 ~ 0.23 m/s, and acceleration a ~ 0.116 m/s².


One needs to understand how to apply the formulas, not just how to plug numbers into equations.

The real world does not have answers in the back of a book. The real world requires hard work and effort.

 

[MoreZitiPlease]

Well, thank you sir.

I love you.

Something tells me you will be seeing me tomorrow.