Answer:
Part a)
![d = 25 km](https://img.qammunity.org/2020/formulas/physics/high-school/gvjhiouw0bcwvwmv6aw0r9yi8nvvvtkay7.png)
Part b)
![v_(avg) = 10.42 km/h](https://img.qammunity.org/2020/formulas/physics/high-school/dmv7g0y39r5qv5lt5ul6ot6rmk91zs72zx.png)
Part c)
![v_(avg) = 17.86 km/h](https://img.qammunity.org/2020/formulas/physics/high-school/892wbeq5cwnl869youmy36nohip6zvamqp.png)
Part d)
Displacement for entire trip = 0
Part e)
Average velocity for entire trip will be zero
Step-by-step explanation:
Part a)
Displacement after t = 2.4 hours is the straight line distance between initial and final positions
so we have
![d = 25 km](https://img.qammunity.org/2020/formulas/physics/high-school/gvjhiouw0bcwvwmv6aw0r9yi8nvvvtkay7.png)
Part b)
Average velocity is defined as
![v_(avg) = (displacement)/(time)](https://img.qammunity.org/2020/formulas/physics/middle-school/zcxbjx3vx9b1va2wbk35xo8p1y2gtkf0z2.png)
![v_(avg) = (25 km)/(2.4 h)](https://img.qammunity.org/2020/formulas/physics/high-school/gjglup8cblu606b0wfyldv62nkh76zk7hg.png)
![v_(avg) = 10.42 km/h](https://img.qammunity.org/2020/formulas/physics/high-school/dmv7g0y39r5qv5lt5ul6ot6rmk91zs72zx.png)
Part c)
During his return journey the displacement will be same
![displacement = 25 km](https://img.qammunity.org/2020/formulas/physics/high-school/9tykn59ssqrpe48ay7i0kbp86mmavn5dej.png)
![time = 1.4 h](https://img.qammunity.org/2020/formulas/physics/high-school/6nvijdlswhjj8qto3qzdeh5geqjefbifc1.png)
so average velocity is defined as
![v_(avg) = (25 km)/(1.4 h)](https://img.qammunity.org/2020/formulas/physics/high-school/bt009npxtc41g3jk4glrbhckzc4l1ro1x3.png)
![v_(avg) = 17.86 km/h](https://img.qammunity.org/2020/formulas/physics/high-school/892wbeq5cwnl869youmy36nohip6zvamqp.png)
Part d)
Displacement for entire trip = 0
as initial and final position will be same
Part e)
Average velocity for entire trip will be zero