From home to office : Time at which she leaves home for the office$= 9 \text{ am}$. speed $=5 \text{ km} \text{ h}^{-1} ;$ distance $=2.5 \text{ km}$ Therefore, time taken to reach the office, $t=\frac{\text{distance}}{\text{speed}}$ $= \frac{2.5 \text{ km}}{5 \text{ km} \text{ h}^{-1}}$ $=0.5 \text{ h}=30 \text{ min}$ Thus, time at which she reaches her office = 9.30 am. Between 9.30 am to 5 pm, she remains in her office i.e. at a distance of 2.5 km from her home. From office to home : Time at which she leaves her office = 5 pm. Speed $=25 \text{ km} \text{ h}^{-1},$ distance $=2.5 \text{ km}$Therefore, time taken to reach the home, $t=\frac{2.5 \text{ km}}{25 \text{ kmh}^{-1}}$$= \frac{1}{10} \text{ h}=6 \text{ min}$Therefore, time at which she reaches her home = 5.06 pm.Hence, (x-t) graph of her motion will be as shown in figure.