Question

need help on example 4 part 1 and 2 if possible

Answer 1

We will solve as follows:

1. We find the linear acceleration in both ends of 40 & 80 meter radius respectively:

*formula:

$w=\alpha r$

"w" is angular acceleration, "a" is linear acceleration & "r" is the radius.

But we knwo that the vehicle is moving at a constant 50.00 m/s so, the linear acceleration is 0m/s^2.

Thus the relationship is 0 m/s^2 at both points. [Option E]

2. We find the angular acceleration in both ends of 40 & 80 meter radius respectively:

*Formula:

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

"a" is centripetal acceleration, "v" is the linear velocity & "r" is the radius.

So:

$a_{40}=_{}\frac{(50.00m/s^{})^2}{(40.00m)}\Rightarrow a_{40}=62.5m/s^2$

&

$a_{80}=\frac{(50.00m/s)^2}{(80m)}\Rightarrow a_{80}=31.25m/s^2$

Now, we find the linear ratio from A with respect to B:

$r=\frac{a_{80}}{a_{40}}\Rightarrow r=\frac{31.25}{62.5}\Rightarrow r=\frac{1}{2}$

So, the ratio is 1/2. [Option A]