Falling Objects

[vivistat6]

The equations for free fall at the surfaces of Planets X and Y are: Planet X: s = 6.14t2;
Planet Y: s = 2.48t2, where s is in meters, and t is in seconds. How long would it
take an object falling from rest to reach a velocity of 33 km h−1 on each planet?

 

[Steven G]

Please read the posting guidelines.

Please note that t2 means t times 2, while t^2 means t squared.

 

[vivistat6]

Indeed. That was my mistake. It is t^2
Thank you.

 

[topsquark]

Please read the posting guidelines.

Please note that t2 means t times 2, while t^2 means t squared.

What is the formula for free-fall on Earth?

-Dan

 

[vivistat6]

Why would it matter what the free fall is on Earth is? This problem is for planet x and planet y

 

[Steven G]

What is the formula for free-fall on Earth?

-Dan

OK, let's see. I never remember these formulas, except for a=-g, but understanding where they come from yields...
a=-g
v = vo-gt
y = yo - voy - gt^2/2

 

[Deleted member 4993]

Why would it matter what the free fall is on Earth is? This problem is for planet x and planet y

it would not matter directly - that exercise will give you a practical "feel" for the problems in hand.

One way to solve the given problems described in OP would be to

Derive equations relating velocity and time - from the given distance -time relations.

 

[nasi112]

OK, let's see. I never remember these formulas, except for a=-g, but understanding where they come from yields...
a=-g
v = vo-gt
y = yo - voy - gt^2/2

They are not on Earth as OP said. I am thinking of,

\(\displaystyle s = (6.14t^2) \ m/s^2\)
\(\displaystyle v = (12.28t) \ m/s^2\)
\(\displaystyle a = (12.28) \ m/s^2\)

The given velocity is \(\displaystyle v = 33 \ km/h = 9.17 \ m/s\).

\(\displaystyle t = \frac{v}{12.28} = \frac{9.17}{12.28} = 0.7467 \ s = 746.7 \ ms\).

You can do the same process for \(\displaystyle s = (2.48t^2) \ m/s^2\).

 

[Deleted member 4993]

it would not matter directly - that exercise will give you a practical "feel" for the problems in hand.

One way to solve the given problems described in OP would be to

Derive equations relating velocity and time - from the given distance -time relations.

ds/dt = v

for the first planet s = 6.14/1000 * t^2

ds/dt =12.28 * t/1000 → 33 = 12.28 * t → t = 33000/12.28 = ? (be careful with units)

continue......

 

[Steven G]

They are not on Earth as OP said. I am thinking of,

\(\displaystyle s = (6.14t^2) \ m/s^2\)
\(\displaystyle v = (12.28t) \ m/s^2\)
\(\displaystyle a = (12.28) \ m/s^2\)

The given velocity is \(\displaystyle v = 33 \ km/h = 9.17 \ m/s\).

\(\displaystyle t = \frac{v}{12.28} = \frac{9.17}{12.28} = 0.7467 \ s = 746.7 \ ms\).

You can do the same process for \(\displaystyle s = (2.48t^2) \ m/s^2\).

OK, let's see. I never remember these formulas, except for a=-g, but understanding where they come from yields...
a=-g
v = vo-gt
y = yo - voy - gt^2/2

Topsquark asked me what is the formula for free-fall on Earth? What is wrong with my response?

 

[Deleted member 4993]

Topsquark asked me what is the formula for free-fall on Earth? What is wrong with my response?

For free-fall a = +g

 

[Steven G]

SK,
Why a=+g? I would think that since the force of gravity is downwards that a=-g.
Steven

 

[Deleted member 4993]

SK,
Why a=+g? I would think that since the force of gravity is downwards that a=-g.
Steven

In terrestrial dynamics, during free-fall, the positive direction is taken to be towards the center of the earth. In that case the velocity and the acceleration vectors point in the same direction. Of course, that is just a matter of convention. The 3 equations that you wrote are generally treated as scalar equation for this purpose (that is v is actually the speed and not the velocity).