# Simple Uniform Acceleration Problem

• TheKovac
In summary: This is incorrect because it assumes constant acceleration, which was not given in the problem. The average acceleration is not the same as the acceleration at any given point in time. You need to use the correct equations for average velocity and average acceleration, as shown in the solution.

## Homework Statement

A bicycles accelerates from rest, covering 16 metres in 4 seconds. The total mass of the bicycle and its rider it 90kg. What is its average acceleration during this time?

x = ut + 0.5at^2
v=xt
a=vt

## The Attempt at a Solution

v = x/t
v = (16)/(4)
v = 4 m/s

F = mv/t
F = (90)(4)/4
Fnet = 90N

Fnet = ma
90 = 90a
A= 1 m/s^2 <--WRONG! - How?
A=2ms <--- RIGHT

TheKovac said:

## Homework Statement

A bicycles accelerates from rest, covering 16 metres in 4 seconds. The total mass of the bicycle and its rider it 90kg. What is its average acceleration during this time?

x = ut + 0.5at^2
v=xt
a=vt

## The Attempt at a Solution

v = x/t
v = (16)/(4)
v = 4 m/s

F = mv/t
F = (90)(4)/4
Fnet = 90N

Fnet = ma
90 = 90a
A= 1 m/s^2 <--WRONG! - How?
A=2ms <--- RIGHT

Why didn't you just use x = ut + 0.5at^2 ?

TheKovac said:

## Homework Statement

A bicycles accelerates from rest, covering 16 metres in 4 seconds. The total mass of the bicycle and its rider it 90kg. What is its average acceleration during this time?

## Homework Equations

x = ut + 0.5at^2
v=xt
a=vt

Your equations are not quite right here and they are leading to some problems. The second and third equations need to be:

$$v_{\rm average} = \frac{x_f-x_i}{\Delta t}$$

$$a_{\rm average} = \frac{v_f-v_i}{\Delta t}$$

## The Attempt at a Solution

v = x/t
v = (16)/(4)
v = 4 m/s

This quantity you have calculated is the average velocity over the whole time period.

F = mv/t
F = (90)(4)/4
Fnet = 90N

Dividing the average velocity by the time here does not give you the average acceleration; you would need the beginning and ending velocities.

However, if you use your first equation:

$$\Delta x = v_0 t + \frac{1}{2} a t^2$$

you should get the right answer immediately.

You can't use a constant acceleration law - The question never said that the acceleration was constant, it wanted you to find the average acceleration.
Also:
average velocity = displacement / time interval
average acceleration = change in velocity / time interval

16/4 = average velocity

!

I would first check the units in my calculation to ensure they are consistent. In this case, the correct unit for acceleration is m/s^2, not m/s. This means that the bicycle's average acceleration during this time is actually 2 m/s^2, not 1 m/s^2.

I would also double check my calculations to see if there were any errors or mistakes made. It is possible that a small error was made in one of the calculations which led to the incorrect answer.

Additionally, I would consider the physical factors that may have affected the bicycle's acceleration, such as friction and air resistance. These factors may have caused the bicycle to have a slightly lower acceleration than the calculated value.

Finally, I would encourage the student to double check their work and to always pay attention to units when solving problems in physics, as they are crucial in obtaining the correct answer.

## What is simple uniform acceleration?

Simple uniform acceleration is a type of motion where an object moves in a straight line with a constant acceleration. This means that the object's velocity increases or decreases by the same amount in each unit of time.

## What is the equation for simple uniform acceleration?

The equation for simple uniform acceleration is a = (vf - vi) / t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time.

## How do I solve a simple uniform acceleration problem?

To solve a simple uniform acceleration problem, you need to know the initial velocity, final velocity, and time. You can use the equation a = (vf - vi) / t to calculate the acceleration. Then, you can use the equations vf = vi + at and d = vit + 1/2at^2 to find the final velocity and displacement, respectively.

## What are the units for acceleration in a simple uniform acceleration problem?

The units for acceleration in a simple uniform acceleration problem are meters per second squared (m/s^2).

## Can simple uniform acceleration be negative?

Yes, simple uniform acceleration can be negative. A negative acceleration means that the object is slowing down, while a positive acceleration means that the object is speeding up.