Question

I got the answers to a and b, but I need help with c!
a)15.4 s
b)displacement= 14.4m@180 degrees
x= 1.73m@180 degrees
Question in the comments

Answer 1

10. A conveyor belt in a factory moves gadgets from east to west in a straight line at a constant speed of 0.825 m/s.
A “quality-control technician” inspects the gadgets from her stationary position alongside the conveyer belt.
Consider her position the origin of our reference frame. Now suppose a defective gadget enters the factory on
the conveyer belt 12.7 meters to the east of the gadget inspector.
(a) How much time elapses before the bad gadget reaches the inspector?
(b) Suppose the inspector is snoozing and doesn’t notice the bad gadget until an instant 2.10 seconds after it
passes her. At that instant, what is the gadget’s position and overall displacement in the factory?
(c) At the instant she notices the bad gadget, she begins walking with constant velocity of 0.950 m/s to the
west. What distance must she walk to catch up to, and retrieve, the bad gadget?

Answer 2

let's set \(t_0 = 0 s\) at the instant she notices the faulty gadget. She walks after the gadget at speed \(v_c = 0.950~ \text{m/s}\), the gadget on the band moves on with \(v_g = 0.825~\text{m/s}\).

At time \(t\) she will have caught up with the gadget. The time is unknown, what we know however is that the gadget will have moved \[d_g = v_g \cdot t\] and the controller will have moved \[d_c = v_c \cdot t\] which is also \[d_c = d_g + v_g \cdot 2.1~\text{s}\]
We can devide the first two equations. that will eliminate \(t\) and gives us an expression for \(d_g\):
\[d_g = \frac{v_g}{v_c}d_c\]
that expression can be inserted into the third equation and we'll get
\[d_c = \frac{v_g}{v_c}d_c + v_g \cdot 2.1~\text{s}\]
We know \(v_g\) and \(v_c\) and when we solve the equation for \d_c\) we have the answer.
\[ d_c \left( 1 - \frac{v_g}{v_c}\right) = v_g \cdot2.1~\text{s}\]
\[\Rightarrow d_c = \frac{v_g \cdot v_c}{v_c - v_g} \cdot 2.1~\text{s} \]