# A question about equations of motion under constant acceleration

In summary, The two equations for a particle moving under constant acceleration, x = \frac{v + v_0}{2}t and x = v_0t + \frac{1}{2}at^2, can both be used to find the time (t) if given certain variables. However, the second equation gives two possible values for t while the first equation only gives one. This is because the two equations have different unknowns and thus, the number of solutions will vary based on the information given. It is important to consider the physical implications and reject any absurd values of t.
This is a very basic question.
We know for a particle moving under constant acceleration we can use both the equations
$$x = \frac{v + v_0}{2}t$$ and
$$x = v_0t + \frac{1}{2}at^2$$If we want to find $t$ form each of these equations, the first one gives only one value but the second one gives two values of $t$.
So it seems like if the first equation is used one value of $t$ will be lost.

Now, I can see one of these equation is linear and another one is quadratic; but still, both of those two equations were derived from same principles - how come one turns out to be better than the other?

This is a very basic question.
We know for a particle moving under constant acceleration we can use both the equations
$$x = \frac{v + v_0}{2}t$$ and
$$x = v_0t + \frac{1}{2}at^2$$If we want to find $t$ form each of these equations, the first one gives only one value but the second one gives two values of $t$.
So it seems like if the first equation is used one value of $t$ will be lost.

Now, I can see one of these equation is linear and another one is quadratic; but still, both of those two equations were derived from same principles - how come one turns out to be better than the other?

No. The two equations are equally good. One of the ##t##'s you get in the second equation should be rejected because it's absurd.

Getting two solutions does not mean that it's true. Can you ever have a negative t?

The two equations do not have the same unknowns. The number of solutions consistent with your knowledge of a system will vary depending on what you know about the system.

In particular, knowing the sign of the final velocity will eliminate one of the two solutions to the second equation.

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No. The two equations are equally good. One of the ##t##'s you get in the second equation should be rejected because it's absurd.

Getting two solutions does not mean that it's true. Can you ever have a negative t?

Would you like to reconsider your statement? Try this problem.

A ball is thrown vertically upwards with a speed 20ms-1 .What is the time elapsed when the ball is at a height 5m ?

Do you get negative 't' ? Which of the two values of time would you reject ?

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1 person
Tanya Sharma said:
A ball is thrown vertically upwards with a speed 20ms-1. What is the time elapsed when the ball is at a height 5m ?

By the way, I see no matter which equation I use it gives two results for the above problem! And both values are valid.

jbriggs444 said:
The two equations do not have the same unknowns.

Well now I see, if I try to have same unknowns the two equations become same.

$y = \frac{v_0+v}{2}t$ If I want to eliminate v from here by $v = v_0+at$ then $y = \frac{2v_0 + at}{2}t = v_0t + \frac{a}{2}t^2$

Thank you all.

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Tanya Sharma said:
A ball is thrown vertically upwards with a speed 20ms-1 .What is the time elapsed when the ball is at a height 5m ?

Do you get negative 't' ? Which of the two values of time would you reject ?
We can't use the first equation here.
Using the second equation, I get
##t=2-\sqrt{3} \approx 0.28##
##t=2+\sqrt{3} \approx 3.72##

Certainly you will have to reject one t here. The two times I got here are equally correct. It's the time taken when the ball is at 5m above the ground. Obviously, the last 't' is the time taken after the ball got it's peak height and returned to 5m above the ground.

By the way, I see no matter which equation I use it gives two results for the above problem!
You can't use ##s=\frac{1}{2}(u+v)t## here because you don't know the final velocity ##v##. You only know the initial velocity.

Using the second equation, I get
##t=2-\sqrt{3} \approx 0.28##
##t=2+\sqrt{3} \approx 3.72##

Then why were you suggesting to the OP that one of the values of 't' would be negative and should be rejected ?

We can't use the first equation here.

I don't understand why you said so. The first equation can be used by putting v = 20 - 9.8t

Certainly you will have to reject one t here.

Why?

Adjacent's comment is valid only for final positions below the initial position. I.e., throwing a stone from a rooftop and asking when will it hit the ground nets two values for t of which one is negative and needs to be discarded as unphysical.
It's not true in general, as has been already shown.

1 person
I don't understand why you said so. The first equation can be used by putting v = 20 - 9.8t
Yes, you can do it that way but if you do it that way, that equation becomes the second equation. So it's like using the second equation again. The first equation cannot be used directly.

Why?
I thought Tanya was asking about the time taken for the ball to reach 5m,while moving upwards.

Oh, I see, Bandersnatch. Thanks for clearing that up. Sorry guys. :shy:
It's been sometime since I last did these kinematics problems

Well now I see, if I try to have same unknowns the two equations become same.

$y = \frac{v_0+v}{2}t$ If I want to eliminate v from here by $v = v_0+at$ then $y = \frac{2v_0 + at}{2}t = v_0t + \frac{a}{2}t^2$

Thank you all.
hehe, yes exactly. The two equations are the same. So what was your original problem? You seemed to say that $y = v_0t + \frac{a}{2}t^2$ gave two solutions? OK, so you mean, if you know the value of ##y## and ##v_0## and ##a## then there are two possible values for ##t##? Yes, I agree with this. And for the other form of the equation $y = \frac{v_0+v}{2}t$ you were saying there is only one solution. So you mean if you are told ##v## and ##y## and ##v_0## then there is only one possible value for ##t##? I agree with this too.

The reason you get two solutions for one case, and only one solution for the other case is because you are using different information each time. To make it clear, for the first equation, you are using ##y## and ##v_0## and ##a## to find ##t## But in the second equation, you are using ##y## and ##v_0## and ##at+v_0## to find ##t##. (since ##v=at+v_0## and you have used ##v##). So it is clear that in the second case, you have used more information, to restrict the possibilities for ##t##.

## 1. What are the three equations of motion under constant acceleration?

The three equations of motion under constant acceleration are the displacement equation (x = x0 + v0t + 1/2at2), the velocity equation (v = v0 + at), and the acceleration equation (v2 = v02 + 2a(x - x0)).

## 2. How do I determine the acceleration of an object using these equations?

To determine the acceleration of an object, you can use the acceleration equation (v2 = v02 + 2a(x - x0)). Plug in the initial velocity (v0), final velocity (v), and displacement (x-x0) values to solve for the acceleration (a).

## 3. Can these equations be used for objects with non-constant acceleration?

No, these equations can only be used for objects with constant acceleration. For objects with non-constant acceleration, more complex equations and calculations are needed.

## 4. How do these equations relate to the motion of an object on a graph?

The displacement equation (x = x0 + v0t + 1/2at2) represents the slope of the position vs. time graph, the velocity equation (v = v0 + at) represents the slope of the velocity vs. time graph, and the acceleration equation (v2 = v02 + 2a(x - x0)) represents the slope of the acceleration vs. time graph.

## 5. How are these equations derived?

These equations are derived from the fundamental equations of motion (x = x0 + vt and v = v0 + at) using calculus and algebraic manipulations.

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