Simple Uniform Acceleration Problem

[TheKovac]

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

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

 

[rock.freak667]

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

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 ?

 

[alphysicist]

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.

 

[XGPAT]

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

 

[asteriatic]

!

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.