Is this equation involving the acceleration of gravity correct?

In summary, according to these equations, a 9.8 m/s2 acceleration is required to move a body a distance of 1 meter in 1 second.
  • #1
Huzaifa
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Homework Statement
Is this equation correct: $$g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8ms^{-2}$$?
Relevant Equations
##g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8ms^{-2}##
$$\begin{aligned}v=u+at\\ \Rightarrow v=gt\\ \Rightarrow g=\dfrac{v}{t} \cdots (1)\end{aligned}$$
$$\begin{aligned}s=ut+\dfrac{1}{2}at^{2}\\ \Rightarrow s=\dfrac{1}{2}gt^{2}\\ \Rightarrow g=\dfrac{2s}{t^{2}}\cdots (2)\end{aligned}$$
$$\begin{aligned}s=vt-\dfrac{1}{2}at^{2}\\ \Rightarrow s-vt=-\dfrac{1}{2}at^{2}\\ \Rightarrow g=\dfrac{2\left( vt-s\right) }{t^{2}} \cdots (3)\end{aligned}$$
$$\begin{aligned}v^{2}-u^{2}=2as\\ \Rightarrow v^{2}=2as\\ \Rightarrow g=\dfrac{v^{2}}{2s} \cdots (4)\end{aligned}$$
$$g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8\ \mathrm{m s^{-2}}$$
 
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  • #2
What's the question precisely?
 
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  • #3
Huzaifa said:
Homework Statement:: Is this equation correct: $$g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8ms^{-2}$$?
Relevant Equations:: ##g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8ms^{-2}##

Huzaifa said:
$$\begin{aligned}v=u+at\\ \Rightarrow v=gt\\ \Rightarrow g=\dfrac{v}{t} \cdots (1)\end{aligned}$$
$$\begin{aligned}s=ut+\dfrac{1}{2}at^{2}\\ \Rightarrow s=\dfrac{1}{2}gt^{2}\\ \Rightarrow g=\dfrac{2s}{t^{2}}\cdots (2)\end{aligned}$$
$$\begin{aligned}s=vt-\dfrac{1}{2}at^{2}\\ \Rightarrow s-vt=-\dfrac{1}{2}at^{2}\\ \Rightarrow g=\dfrac{2\left( vt-s\right) }{t^{2}} \cdots (3)\end{aligned}$$
$$\begin{aligned}v^{2}-u^{2}=2as\\ \Rightarrow v^{2}=2as\\ \Rightarrow g=\dfrac{v^{2}}{2s} \cdots (4)\end{aligned}$$
$$g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8\ \mathrm{m s^{-2}}$$
What you have is multiple equations.

You need to describe the overall situation. Also define precisely what each variable means.

Of course, answer the question asked by @PeroK. .
 
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  • #4
SammyS said:
You need to describe the overall situation. Also define precisely what each variable means.
The overall situation is constant or uniform acceleration. Here, s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time. These are equations of motion.
 
  • #5
None of the numbered equations [except maybe (3)] are generally true.
Each is true in some special [so far unstated] cases
 
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  • #6
Hi @Huzaifa. You seem to be making a number of unstated assunptions, e.g.
- motion is in the vertical direction only, with gravity the only force;
- initial velocity (u) is zero, so you are only considering objects released from rest at t=0.

But of course I'm just guessing, as you haven't yet answered @PeroK's question (Post #2).
 
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FAQ: Is this equation involving the acceleration of gravity correct?

What is the equation for acceleration of gravity?

The equation for acceleration of gravity is a = g, where a is the acceleration and g is the gravitational constant (approximately 9.8 m/s² on Earth).

How is the acceleration of gravity calculated?

The acceleration of gravity is calculated by dividing the force of gravity on an object by its mass. This can be represented by the equation a = F/m, where F is the force of gravity and m is the mass of the object.

Is the acceleration of gravity constant?

Yes, the acceleration of gravity is considered to be constant in a vacuum. However, it can vary slightly depending on factors such as altitude and latitude on Earth.

What are the units for acceleration of gravity?

The units for acceleration of gravity are typically meters per second squared (m/s²) or feet per second squared (ft/s²).

Can the acceleration of gravity be negative?

Yes, the acceleration of gravity can be negative if the object is moving in the opposite direction of the gravitational force. This is often seen in free fall or when an object is thrown upwards.

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