How would a graph of acceleration vs 1/mass look like?

Question

Fnet = ma
Fnet/a =m
a/Fnet = 1/m
a / (1/m) = Fnet
so the slope of the graph is Fnet

but I can't seem to figure out what it would look like, so i tried to convert it into slope, x , y:
slope = m , x = m , y = a

normally:
m = xy
m/x = y

a vs 1/m:
y/(1/x) = m
y = m*(1/x)
y = m/x

so I graphed it on an online calculator and i get a exponential function, am I right because some of my classmates insists its a linear function/ graph and the teacher told him he's right....but she told me i was right too....?

Answer 1

You have made a mistake. You wrote:

"normally:

m = xy

m/x = y"



That's wrong. You mean:

m = y/x

y = mx

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The question is best stated as "How would a graph of acceleration vs 1/mass for a constant force look?"



Rahul's answer is short and clear. But to help, let me give an example with numbers.

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The easiest way to see what happens is to make up some simple values. For example:



Suppose F = 10N. This remains fixed.



F = ma. So a = F/m. We can write this as: a = F * (1/m)



If m=10kg

1/m = 1/10 = 0.1 kg^-1

a = F * (1/m)= 10 * 0.1 = 1m/^2



If m=5kg

1/m = 1/5 = 0.2 kg^-1

a = F * (1/m) = 10 *0.2 = 2m/^2



If m=1kg

1/m = 1/1 =1.0 kg^-1

a = F * (1/m) = 10 * 1.0 = 10m/^2



So using '1/m' values on the x axis and 'a' values on the y axis, the points will be:

(0.1, 1)

(0.2, 2)

(1.0, 10)

which is a straight line passing through the origin. The gradient is 10 (which is the force).

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Here is a more detailed explanation:



F = ma, so a = F/m. Write this as:

a = F * (1/m) (equation 1)



You should see that because F is a constant, equation 1 is really the same as Y=MX (I've used capitals as we don't want to confuse M (gradient) with m (mass).)



Compare a = F * (1/m) to Y=M * X

a corresponds to the variable Y.

F corresponds to the constant M (which is the gradient)

(1/m) corresponds to the variable X.



So a graph of a vs. (1/m) will have gradient equal to F.

Answer 2

The graph will be a straight line with the slope of Fnet. It is same as y = mx where x is 1/mass and m is Fnet and y is a

Answer 3

it would depend on what the velocity graph and/or the distance vs. time graph would look like.



if the distance graph is exponential

the velocity graph would be linear

and the acceleration graph would be a straight horizontal line.



they are all derivatives of each other