Calculating Distance and Time for an Object on a Frictionless Inclined Plane

[physicsnewby]

Homework Statement

1 An object is pushed up a smooth incline of 30 degrees with a speed of 4m/s. In seconds, how long will it take before the object stops and reverses direction?

Vf= Vo - at

Vf= Vo - at
0 = 4 - (9.8)t
t = 0.41s

I know this is wrong because the angle isn't accounted for, but I'm not sure where to use it. I'm not sure what the right answer is either.
2 An object is placed on an incline of 30 degrees . The object is pushed up the incline at the initial speed vo=4m/s. How far does it travel before stopping and reversing direction?

Vf= Vo - at

Vf= Vo - at
0 = 4 - (9.8)t
t = 0.41s

x= (Vxo)(t)
x= (4)(0.41)
x=1.6m

For some reason, this gives me the right answer, but I again don't use the 30 degrees anywhere. What should I be doing?

 

[Doc Al]

Vf= Vo - at

Vf= Vo - at
0 = 4 - (9.8)t
t = 0.41s

This is incorrect. First find the actual acceleration by considering the component of gravity acting along the incline. (This will involve a bit of trig.)

x= (Vxo)(t)
x= (4)(0.41)
x=1.6m

This is incorrect because the speed is not constant. You'd need the average speed to find the distance using this method.

For some reason, this gives me the right answer, but I again don't use the 30 degrees anywhere.

Amusingly, the two errors you made exactly cancel.

 

[physicsnewby]

ok, then to get acceleration, would I use:

a = g sin theta
a = 9.8 sin 30
a = 4.9 m/s^2

then to solve for time:
Vf= Vo - at
0 = 4 - (4.9)t
t = 0.82 s

 

[Doc Al]

ok, then to get acceleration, would I use:

a = g sin theta
a = 9.8 sin 30
a = 4.9 m/s^2

then to solve for time:
Vf= Vo - at
0 = 4 - (4.9)t
t = 0.82 s

So far, so good.

 

[physicsnewby]

Question 1 is then answered.

Question 2, if I need avg. speed:

V avg = Vf + Vo / 2
V avg = 0 + 4 / 2
V avg = 2

X = Xo + volt + 1/2at^2
X = 0 + (2)(0.82) +1/2(o)^2
X = 1.64m

 

[Doc Al]

Question 2, if I need avg. speed:

V avg = Vf + Vo / 2
V avg = 0 + 4 / 2
V avg = 2

That's the correct average speed. To get the distance use V_{ave} = \Delta X / \Delta T.

X = Xo + volt + 1/2at^2
X = 0 + (2)(0.82) +1/2(o)^2
X = 1.64m

Using this equation is a perfectly fine way to solve for distance, but you need to do it right. Two problems: (1) Vo is the initial velocity, not average velocity; (2) What happened to the last term? (1/2at^2) It's not zero!

 

[physicsnewby]

For Question 2 using the formula you gave, then

Vavg = delta X/ delta t
2 = delta X / 0.82
delta X = 1.64m

I got the same answer, but my original equation was all wrong.
X = 0 + (2)(0.82) +1/2(o)^2

with correct numbers
X = 0 + (4)(0.82) +1/2(4.9)^2
X = 3.28 + 12.01
X = 15.29m

Which one is right?

 

[Doc Al]

For Question 2 using the formula you gave, then

Vavg = delta X/ delta t
2 = delta X / 0.82
delta X = 1.64m

Correct method = correct answer.

I got the same answer, but my original equation was all wrong.
X = 0 + (2)(0.82) +1/2(o)^2

Wrong values = wrong answer.

with correct numbers
X = 0 + (4)(0.82) +1/2(4.9)^2
X = 3.28 + 12.01
X = 15.29m

The last term (1/2at^2) is not correct. What is the acceleration? (Careful with signs--if up the incline is positive, then down the incline must be negative.) What is the time?

This equation for uniformly accelerated motion will come up often, so be sure you get it straight.

 

[physicsnewby]

I've often had problems with these questions. When I put a value in for acceleration, my teacher always circles it and says there should be no acceleration when solving for distance X, so I get confused a lot!

I know I've got the right answer, but want to know how to correct this other equation.

X = 0 + (4)(0.82) +1/2(4.9)(0.82)^2
X = 3.28 - 1.64
X = 1.64m !

THANKS !

 

[Chronos]

cosine theta of the inclination [in a gravitational field] would be worthy of consideration.

 

[Doc Al]

I've often had problems with these questions. When I put a value in for acceleration, my teacher always circles it and says there should be no acceleration when solving for distance X, so I get confused a lot!

Not all motion is accelerated. For example, in projectile motion problems the acceleration only affects the vertical motion, not the horizontal. In such a problem there will be no acceleration when solving for horizontal (x-axis) distance.

I know I've got the right answer, but want to know how to correct this other equation.

X = 0 + (4)(0.82) +1/2(4.9)(0.82)^2
X = 3.28 - 1.64
X = 1.64m !

The only error that I see is having the wrong sign for the acceleration in the first line. The magnitude of the acceleration down the incline will be g \sin 30 = g/2 = 4.9, but the sign should be negative, since it acts down the incline.

 

[Chronos]

In a projectile problem, you have two forces - gravity and propulsion.

 

[Doc Al]

In a projectile problem, you have two forces - gravity and propulsion.

Once an object has been thrown or released, it becomes a projectile. Ignoring air resistance, the only force acting on it is gravity. (Once released, any propulsive force no longer acts.)

 

[Chronos]

Correction noted. Initial velocity is the proper terminology.