# Simple acceleration problems - requesting a check

• RyanJF
In summary, the conversation involved a person seeking help with their homework problems, specifically for problem number three where they were unsure of the equation to use. The problems involved calculating speed and time using given equations and variables. The person eventually figured out the correct equation for problem number three and arrived at an answer of 17.5 meters.
RyanJF
This is a bit of reading, but the math is simple. I only need my work checked, except for problem number three, in which case I'm simply asking for clarification on which equation I should use.

## Homework Statement

Problem number one:

"A flower pot falls from a windowsill 25m above the sidewalk. A) How fast is the flower pot moving when it strikes the ground? B) How much time does a passerby on the sidewalk below have to move out of the way of the way before the flower pot hits the ground?"

a = 9.8 (gravity)
x = 25
Vi = 0

Problem number two:

"A plane, starting from rest at one end of a runway, undergoes a constant acceleration of 1.6 m/s^2 for a distance of 1600 meters before takeoff. A) What is its speed upon takeoff? B) What is the time required for takeoff?"

a = 1.6
Vi = 0
x = 1600

Problem number three:

"A boy sliding down a hill accelerates at 1.4m/s^2. If he started from rest, in what distance would he reach a speed of 7/ms?"

a = 1.4
Vi = 0
Vf = 7

## Homework Equations

Problem number one:

v^2 = Vi^2 + 2ax

Problem number two:

v^2 = Vi^2 + 2ax

Problem number three:

v = Vi + at
x = 1/2 (Vi + V)t
x = Vi x T + 1/2at^2
v^2 = Vi^2 + 2ax

## The Attempt at a Solution

Problem number one:

v^2 = Vi^2 + 2ax

Plugging in numbers for the variables gives me

v^2 = 0^2 + 2 x 9.8 x 25

so

v^2 = 490

sq(490 = 22.135

V = 22.135 m/s

Part b, all I did was:

25/22.135 = 1.129 seconds for the passerby to move out of the way of the falling flower pot.

Problem number two:

v^2 = Vi^2 + 2ax

Plugging in the numbers for the variables gives me

v^2 = 0^2 + 2 x 1.6 x 1600

so

v^2 = 5120

sq(5120

V = 71.554 m/s

For part B, all I did was:

t = Vf - Vi / A

Plugging in the numbers for the variables gives me

t = 71.554 - 0 / 1.6

so

t = 44.72 seconds for the plane to take off

Problem number three:

Not sure of the equation, so I didn't start it.

===

If I didn't already say, all I'd like is to have my answers for problems one and two checked, with an equation(s) supplied for problem number three.

===

Hopefully somebody will help me out here. Thanks!

Eh, I think the answer to problem three is 17.5 meters. All I had to do was take another of the equations and solve for T, which is what I originally thought. Answer makes sense.

I need to stop being lazy.

## 1. What is simple acceleration and how is it different from regular acceleration?

Simple acceleration refers to the rate at which an object's velocity changes over time. It is different from regular acceleration in that it only takes into account the initial and final velocities of an object, while regular acceleration also considers any changes in direction.

## 2. How do I calculate acceleration in a simple acceleration problem?

To calculate acceleration in a simple acceleration problem, you can use the formula: a = (vf - vi) / t, where a is acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

## 3. What units are used to measure acceleration?

Acceleration can be measured in a variety of units depending on the system of measurement being used. In the metric system, it is typically measured in meters per second squared (m/s²). In the imperial system, it is often measured in feet per second squared (ft/s²).

## 4. How can I check if my answer to a simple acceleration problem is correct?

You can check your answer by plugging in the values you calculated into the original formula and making sure it equals the correct answer. You can also double-check your calculations and units to make sure they are correct.

## 5. Can I use simple acceleration to solve problems involving gravity?

Yes, simple acceleration can be used to solve problems involving gravity, as gravity is a force that causes objects to accelerate towards the Earth. However, in these cases, it is important to also consider the acceleration due to gravity (9.8 m/s²) in your calculations.

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