Average Vector Acceleration Along a Curve and a Straight Section

[srekai]

Homework Statement

f

f

Homework Equations

$a = {v^2}{r}$
$d = vt$

The Attempt at a Solution

The way I do it is break it down into the curved and straight segment.
The bicycle is always traveling at a constant speed.
Curved portion:
distance $\frac{\pi R^2}{4}$
acceleration $= \frac{v^2}{R}$
time $t = \frac{\frac{\pi R^2}{4}}{v}$

Straight portion:
distance $R$
acceleration 0
time $t = \frac{R}{v}$

Combining these two pieces
avg acc. = $\frac{\frac{v^2}{R} \cdot \frac{\pi R^2}{v}}{\frac{\pi R^2}{v} +\frac{R}{v}} = \frac{\pi Rv}{\frac{\pi R^2+R}{v}} = \frac{\pi v^2}{\pi R + 1}$

As for the direction, I'm assuming it's to the right?

 

[FactChecker]

Since the acceleration around the arc is always pointing at the center of the circle, what would be the average direction on that part? Then there is no acceleration in the straight part, so the direction is not changed.

PS. When you use the term "right" you should be clear if you are talking about the right of the diagram or to the right of the velocity vector forward direction.

 

[haruspex]

$a = {v^2}{r}$

I assume you meant v²/r. That is the magnitude of the acceleration. You are told to average acceleration as a vector.
What is the definition of average acceleration?

distance $\frac{\pi R^2}{4}$

That is an area, not a distance.

 

[PeterO]

Homework Statement

fView attachment 229300
f

Homework Equations

$a = {v^2}{r}$
$d = vt$

The Attempt at a Solution

The way I do it is break it down into the curved and straight segment.
The bicycle is always traveling at a constant speed.
Curved portion:
distance $\frac{\pi R^2}{4}$
acceleration $= \frac{v^2}{R}$
time $t = \frac{\frac{\pi R^2}{4}}{v}$

Straight portion:
distance $R$
acceleration 0
time $t = \frac{R}{v}$

Combining these two pieces
avg acc. = $\frac{\frac{v^2}{R} \cdot \frac{\pi R^2}{v}}{\frac{\pi R^2}{v} +\frac{R}{v}} = \frac{\pi Rv}{\frac{\pi R^2+R}{v}} = \frac{\pi v^2}{\pi R + 1}$

As for the direction, I'm assuming it's to th

Homework Statement

fView attachment 229300
f

Homework Equations

$a = {v^2}{r}$
$d = vt$

The Attempt at a Solution

The way I do it is break it down into the curved and straight segment.
The bicycle is always traveling at a constant speed.
Curved portion:
distance $\frac{\pi R^2}{4}$
acceleration $= \frac{v^2}{R}$
time $t = \frac{\frac{\pi R^2}{4}}{v}$

Straight portion:
distance $R$
acceleration 0
time $t = \frac{R}{v}$

Combining these two pieces
avg acc. = $\frac{\frac{v^2}{R} \cdot \frac{\pi R^2}{v}}{\frac{\pi R^2}{v} +\frac{R}{v}} = \frac{\pi Rv}{\frac{\pi R^2+R}{v}} = \frac{\pi v^2}{\pi R + 1}$

As for the direction, I'm assuming it's to the right?

Surely Average Acceleration is simply (final Velocity - initial velocity)/time taken.
In this case the subtraction is the subtraction of two vectors, and you have to work out an expression for the time interval, using speed and distance.

 

[Chestermiller]

Surely Average Acceleration is simply (final Velocity - initial velocity)/time taken.
In this case the subtraction is the subtraction of two vectors, and you have to work out an expression for the time interval, using speed and distance.

I think that this is what @haruspex was (more subtly) alluding to.