Velocity-distance graph for a freely falling body

[g9WfI]

Homework Statement

Which graph, A, B, C or D, best represents the relationship between the variables x and y where:
y is the velocity of an object falling freely from rest, x is the distance fallen?

I attach choices of graphs below.

Relevant Equations

Equations of motion

My reasoning:
Gradient of graph = v/s = t
t increasing therefore gradient increasing
So graph B

The answer is C.

 

[BvU]

Hi,

v/s = t is only true if v is constant. In free fall v is not constant.

If you write down equations for x and y as a function of time you can eliminate t to find v(s) .

$\$

 

[haruspex]

v/s = t

If you divide metres per second by metres, what do you get?

v/s = t is only true if v is constant

Not even then.

 

[BvU]

Oops ! Slight oversight . But even s = v t does not apply.

Thankie Australia !

$\$

 

[g9WfI]

If you divide metres per second by metres, what do you get?

Ah - 1/t
So as t increases, gradient decreases.
So answer is C.

 

[g9WfI]

If you write down equations for x and y as a function of time you can eliminate t to find v(s) .

I get v = 9.8 x √ (s/4.9)
Putting that into a graphing calculator I get a graph which looks like C, but I wouldn't have been able to recognise that myself :/

 

[PeroK]

I get v = 9.8 x √ (s/4.9)
Putting that into a graphing calculator I get a graph which looks like C, but I wouldn't have been able to recognise that myself :/

It makes sense if you think about energy. For a given distance, the loss in GPE is constant: in this case $mg\Delta x$.

But, the increase in KE is $\frac 1 2 m[(v + \Delta v)^2 - v^2] = \frac 1 2 m\Delta v(2v + \Delta v)$.

As $x$ increases, $v$ increases and the change $\frac{\Delta v}{\Delta x}$ must decrease.

 

[haruspex]

Ah - 1/t
So as t increases, gradient decreases.
So answer is C.

As @BvU notes, v/s=1/t is only true for constant velocity. More generally, $v_{avg}/s=1/t$, but that is with v averaged over time. There is no easy way to relate that to the instantaneous v used in the graph axis.

I get v = 9.8 x √ (s/4.9)
Putting that into a graphing calculator I get a graph which looks like C, but I wouldn't have been able to recognise that myself :/

If you write it instead as $v^2=2gs$ you should recognise that as a parabola. Setting y=2gs and x=v it gives the classic $y=x^2$, so you should be looking for that graph but flipped around, swapping the x and y axes.