3. A car travels along a straight line at a constant speed of `60.0 mi//h` for a distance `d` and then another distance `d` in the same direction at another constant speed. The average velocity for the entire trip is `30.0 mi//h`. (a) What is the constant speed with which the car moved during the second distance `d`? (b) What If? Suppose the second distance `d` were traveled in the opposite direction; you forgot something and had to return home at the same constant speed as found in part (a). What is the average velocity for this trip? (c) What is the average speed for this new trip?
`v = d/(t)`
(a) `20.0\ mi//h` (b) `0` (c) `20.0\ mi//h`
Given: `v_1 = 60.0mi//h` `overline(v) = 30.0mi/h` Calculations: (a) `∆t_1 = d/v_1` `∆t_2 = d/v_2` `overline(v) = (2d)/(∆t_1 + ∆t_2) = (2d)/(d/v_1 +d/v_2)=(2)/(1/v_1 +1/v_2)` `=>overline(v)(1/v_1 +1/v_2) = 2` `=>overline(v)((v_1 +v_2)/(v_1v_2)) = 2` `=>overline(v)(v_1 +v_2) = 2(v_1v_2)` `=>overline(v)v_1 +overline(v)v_2 = 2v_1v_2` `=>overline(v)v_1 = 2v_1v_2 -overline(v)v_2` `=>overline(v)v_1 = (2v_1 -overline(v))v_2` `=>v_2= (overline(v)v_1)/(2v_1 -overline(v))` `=>v_2 = (30.0 mi//h times 60.0\ mi//h)/(2(60.0mi//h)-30.0mi/h)`
`v_2= color(red)(20.0mi//h)`(b) Since the car back to the initial position, `∆x = 0` so velocity `v = (∆x)/(∆t) = color(red)(0)` (c) Since speed deals with distances covered in time intervals. SO the average speed is the same as the average velocity. `v= color(red)(20.0 mi//h)`