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=(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)=((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=(a_(80))/(a_(40))\Rightarrow r=(31.25)/(62.5)\Rightarrow r=(1)/(2)`

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