# Motion with constant acceleration

• SpaceW
In summary, the conversation discusses a problem involving a person dropping a rock into a well and trying to calculate the depth of the well using the speed of sound and equations for constant and accelerated motion. The attempted solution involves setting up an equation and solving for the depth, but the answer obtained is incorrect, leading to a discussion about possible mistakes made in the calculations.
SpaceW

## Homework Statement

A person is standing over a well and drops a rock. After 2.4 seconds he hears a rock splashing into the water. Take the speed of sound to be 300m/s and calculate the depth of the well.

## Homework Equations

x=v/t for constant velocity
x=1/2at^2 for accelerated motion

## The Attempt at a Solution

I tried this solving by taking t1=x/v x is depth of well and v speed of sound and also t2=(2x/g)^1/2
t1+t2=2.4 but i keep getting x to be 104.56 meters which can't be right although it seems all my steps are conclusive. Substituted t1 and t2 took p=x^1/2 and solved quadratic equation for p and god 10.4 m which goes to 104 m = x which is not right. What do you think is wrong?

As a check work the numbers backwards..

On the way down..

d = ut+ 0.5gt2
u=0 so..
t = SQRT(2d/g) = SQRT(2*104/9.81) = 4.6 seconds

So you made a mistake somewhere.

T1+T2=2.4
X/v + ((2x)^1/2)/g^1:2 = 2.4
(g^1/2)x + v(2^1/2)x^1/2 = c
Call c = 2.4vg^1/2
take p=x^1/2
(g^1/2)p^2 + v(2^1/2)p - c = 0
I get p = 10.4 m

SpaceW said:
T1+T2=2.4
X/v + ((2x)^1/2)/g^1:2 = 2.4
(g^1/2)x + v(2^1/2)x^1/2 = c
Call c = 2.4vg^1/2
take p=x^1/2
(g^1/2)p^2 + v(2^1/2)p - c = 0 [*]
I get p = 10.4 m

I think your equation [*] is correct, but you made a mistake solving it.

I solved it twice getting the same answer

SpaceW said:
I solved it twice getting the same answer

Maybe you made the same mistake twice. You didn't forget something in the denominator of the formula for quadratic roots by any chance?

SpaceW
Oops 2a, not a...

## 1. What is constant acceleration?

Constant acceleration is when an object's velocity changes at a constant rate over time. This means that the object's speed and direction are changing at a consistent rate.

## 2. How is constant acceleration calculated?

Constant acceleration can be calculated using the formula a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time elapsed.

## 3. What is the difference between constant acceleration and uniform motion?

Constant acceleration refers to a change in velocity over time, while uniform motion refers to a constant velocity over time. In other words, constant acceleration involves a change in speed or direction, while uniform motion does not.

## 4. What is the relationship between distance and time in motion with constant acceleration?

When an object is experiencing constant acceleration, the distance it travels is directly proportional to the square of the time elapsed. In other words, if the time is doubled, the distance traveled will be four times the original distance.

## 5. What are some real-life examples of motion with constant acceleration?

Some examples of motion with constant acceleration include a car accelerating from a stop, a falling object under the influence of gravity, and a train gradually picking up speed.

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