Calling all Physics People, help!

[johnjohn320]

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0 degrees below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 4.00m/s^2 for a distance of 50.0 m to the edge of the cliff. The cliff is 30.0m above the ocean. Find (a) the car's position relative to the base of the cliff when the car lands in the ocean, and (b) the length of time the car is in the air.

Now, we just started this unit today, and all the problems we did in class were easy because you just separated the horizontal direction from the vertical direction of motion, and could solve for either one easily. But this problem throws a few new things at me. Basically, I could figure this problem out fine if I knew two things: How do I split the acceleration of 4 between the horizontal and verical directions, and what on earth does the 24 degree angle have to do with anything?

If someone could help me out with those two things, I think I can solve the problem just fine.

Thanks!

 

[Brutuskend]

click me

 

[dmurray14]

dude, that's a pretty fux0red question.

or something.

 

[dighn]

the wording is a little confusing

 

[johnjohn320]

Originally posted by: dmurray14
dude, that's a pretty fux0red question.

or something.

Geez, at least Brutuskend's useless reply was funny...

 

[crumpet19]

For some reason I'm compelled to say that you must still separate the hozir and vert components. But, its been a few semesters since I've taken any physics. Try using COS and SIN.

 

[Brutuskend]

Originally posted by: johnjohn320

Originally posted by: dmurray14
dude, that's a pretty fux0red question.

or something.

Geez, at least Brutuskend's useless reply was funny...

I aims to please!

 

[dmurray14]

Originally posted by: johnjohn320

Originally posted by: dmurray14
dude, that's a pretty fux0red question.

or something.

Geez, at least Brutuskend's useless reply was funny...

hmm, well let me at least try to be a bit helpful....


the 24 degree thing is part of the problem because somehow you need to figure out how fast the car has accelerated to due to the 24 degree slope that is 50m long...how, i don't know, sorry.

 

[MikeMike]

you dont split it.

it will have an acceleration of 9.8m/s^2 once it leaves the cliff.

ill give u 4 equations, u figure out which one to use. (th symbols should be self explanatory)

Vf=Vi+at
d=1/2(Vf+Vi)t
d=Vit+1/2at^2
Vf^2=Vi^2 + 2ad

Vf=velocity final
Vi=initial velocity
t=time
a=accel
d=distance
t=time

try to work from there.

something about the 24degrees seems to me like you need to figure out how high the cliff is above the ocean after it falls for 24degrees for 50m.


MIKE

 

[RaynorWolfcastle]

I'm too lazy to work it out but, the 24 deg. angle is used so that you can find the vector of his initial velocity. Also use trig to solve the hroizontal and vertical components theta = arctan(vertical/horiz). Pretty straight forward problem after that, you just have to plug into a couple of the equations of motion

 

[dullard]

That is simple.

1) You know the acceleration (4 m/s^2) and you know the distance (50 m). Thus you can calculate the velocity it will be at the end of the cliff. You do this without any regard to the angle. This will be exactly what you did in previous chapters.
2) Take this velocity and split it into horizontal and vertical velocities (use the sin and cos functions with 24°).
3) Do the simple problems you did in the past for the veritical direction (but now use the 9.8 m/s^2 acceleration of gravity). This will tell you how long the car is in the air.
4) Take the time in the air and multiply it by the horizontal velocity from part (2). This will tell you how far the car traveled.

The only thing confusing I see is that questions (a) and (b) must be answered in the opposite order than asked.

 

[Evadman]

I won't do your homework for you, but I will help you.

#1 Find the horizontal and vertical speeds of the car as it falls off the cliff. This can be done much easier since your teacher has already solved for gravity for you by giving you the 4.00 m/s^2.

#2. Find out how long the car was in the air. This is done by taking the vertical speed the car has, and then changing from 4.00 m/s^2 to 9.81m/s^2. This will give you a a final speed, and the time the car was in the air.

#3. Since the teacher has already given you the 4.00 m/s^2 accel, then you do not need the 24 degrees for jack squat. It would be exactly the same as if the car was on flat level ground accelerating at 4.00M/s 30m up.

<Edit> oh wait, you do need the 24 degrees in #1. my head is not working today.

 

[DrPizza]

Heyyyyyyy, I like that question! (I like giving PITA problems on occasion)

1st, find the velocity of the car at the edge of the cliff... (same problem as acceleration of 4.0m/s/s on a horizontal surface for 50.0 meters)

2nd, break that velocity into the horizontal and vertical components...

3rd, I'd solve for time in the air. s=Vit + 1/2 at^2 Use the y-component of velocity at the edge of the cliff for Vi, s = 30.0 meters, and a = 9.81 m/s/s (I defined down as the positive direction) Get s on the other side of the equation, 1/2at^2 + Vit - 30 = 0. Find the roots of the equation; use the quadratic formula with a=1/2(9.81), b=Vi, and c=-30. Solve for t; you'll get a positive and a negative answer... reject the negative (obviously)

4th: Using the time obtained in step 3, and the x-component of velocity at the edge of the cliff, find how far it goes from the edge of the cliff.

 

[johnjohn320]

Originally posted by: nourdmrolNMT1
you dont split it.

it will have an acceleration of 9.8m/s^2 once it leaves the cliff.

ill give u 4 equations, u figure out which one to use. (th symbols should be self explanatory)

Vf=Vi+at
d=1/2(Vf+Vi)t
d=Vit+1/2at^2
Vf^2=Vi^2 + 2ad

Vf=velocity final
Vi=initial velocity
t=time
a=accel
d=distance
t=time

try to work from there.

something about the 24degrees seems to me like you need to figure out how high the cliff is above the ocean after it falls for 24degrees for 50m.


MIKE

OK, so the angle of 24 degrees has nothing to do with anything then?

I solved vf^2=vi^2+2ad for the time to the edge of the cliff, to get the initial velocity of the car as it left the cliff (I got 20m/s).

So now I'm on the time the car is in the air. I have a vi (20) and an acceleration (-9.8). Now I have the given distance of the 30 m, but that's not the distance the car is going to travel (since it doesnt go immediately straight down). This is where I get confused.

 

[dighn]

no you need the 24 degrees to break down the initial velocity into the x and y components . the verticle component can be used as the initial veclotiy in 1/2at^2 + Vit

 

[DrPizza]

p.s. I'm a physics and calculus teacher.... I'd go with my solution, rather than dullard's

 

[johnjohn320]

Originally posted by: DrPizza
Heyyyyyyy, I like that question! (I like giving PITA problems on occasion)

1st, find the velocity of the car at the edge of the cliff... (same problem as acceleration of 4.0m/s/s on a horizontal surface for 50.0 meters)

2nd, break that velocity into the horizontal and vertical components...

3rd, I'd solve for time in the air. s=Vit + 1/2 at^2 Use the y-component of velocity at the edge of the cliff for Vi, s = 30.0 meters, and a = 9.81 m/s/s (I defined down as the positive direction) Get s on the other side of the equation, 1/2at^2 + Vit - 30 = 0. Find the roots of the equation; use the quadratic formula with a=1/2(9.81), b=Vi, and c=-30. Solve for t; you'll get a positive and a negative answer... reject the negative (obviously)

4th: Using the time obtained in step 3, and the x-component of velocity at the edge of the cliff, find how far it goes from the edge of the cliff.

That is the step I don't get it. People are talking about the sin and cos functions, and while I'm familiar with trig, I swear it was never mentioned in Physics class today. I have no idea how to split the velocity between horizontal and vertical directions. Help DrPizza!

 

[kevinthenerd]

http://honors.saintleo.edu/durette/files/car_problem.gif

 

[dullard]

Originally posted by: DrPizza
p.s. I'm a physics and calculus teacher.... I'd go with my solution, rather than dullard's

We always seem to post in the same threads. Now was that a joke (hence the smiley) or did I do something stupid that I cannot see?

 

[MikeMike]

he wants it relative to cliff? or displacement (unless u havent gotten here yet, i juat got to displacement of 2d graphs, its pretty easy

MIKE

 

[her209]

d=s0+v0t+.5at^2

50=0+0+.5(4)t^2

100 = 4t^2

25 = t^2

t=5

deriving d by t, we get

d/t=v=v0+at

v=0+4*5=20

so the car is moving 20 m/s when it hits the end of the incline

you then need to use the sin rule to find the horizontal & vertical velocity component

vx = 20 sin (90-24) ~ 18.27 m/s
vy = 20 cos (90-24) ~ 8.13 m/s

Next you need to figure out how long it will take for the car to drop 30 m starting at the initial velocity of 8.13 m/s. (Part B)

Then you can take that time and plug it into the d=vxt eq to figure out how far away it is from the edge of the cliff. (Part A)

 

[MrPickins]

OK, so the angle of 24 degrees has nothing to do with anything then?

Originally posted by: RaynorWolfcastle
... the 24 deg. angle is used so that you can find the vector of his initial velocity. Also use trig to solve the hroizontal and vertical components theta = arctan(vertical/horiz)...

Edit: should have clicked refresh before posting...

 

[Rayden]

this is what i studied last week.

the initial velocity is at the angle given in the problem. not horizontal velocity. This means the horizontal velocity and the vertical velocitywill add up to the total velocity at that angle.
you use trigonometry (sins and cosines) to figure out the horizontal and vertical velocities.

then you use the 4 equations for constant acceleration problems using your initial NEGATIVE vertical velocity as the initial velocity and acceleration as -9.80 m/s^2 solving for time.

then Displacement (distance) = time * horizontal velocity


how far the car rolled (50m) before leaving the cliff seems to be the only unimportant piece of data. no wait. i guess you need that along with 4.00m'/s^2 acceleration. i was thinking that was the velocity.
once you do a few of these then they are easy.

 

[RaynorWolfcastle]

Originally posted by: johnjohn320
That is the step I don't get it. People are talking about the sin and cos functions, and while I'm familiar with trig, I swear it was never mentioned in Physics class today. I have no idea how to split the velocity.

Damn, I'll help since this thread will stay up here all night if someone doesn't do it.

you have a vector in polar form (4m/s^2@-24deg) now you know that sin(theta) = opposite/hypotenuse and cos(theta) = adjacent/hypotenuse, so:
-4*sin(24) = vertical component
4*cos(24) = horiz component

(I used a couple of simple trig identities to further simplify this for you)

 

[johnjohn320]

Originally posted by: her209
d=s0+v0t+.5at^2

50=0+0+.5(4)t^2

100 = 4t^2

25 = t^2

t=5

deriving d by t, we get

d/t=v=v0+at

v=0+4*5=20

so the car is moving 20 m/s when it hits the end of the incline

you then need to use the sin rule to find the horizontal & vertical velocity component

vx = 20 sin (90-24) ~ 18.27 m/s
vy = 20 cos (90-24) ~ 8.13 m/s

Next you need to figure out how long it will take for the car to drop 30 m starting at the initial velocity of 8.13 m/s. (Part B)

Then you can take that time and plug it into the d=vxt eq to figure out how far away it is from the edge of the cliff. (Part A)

Ok, I think that was the part I was missing. Had never seen those functions before. OK, I think I might have it now, thanks!