91Ó°ÊÓ

The total distance between the two towns is, 300 km

The speed for the first 100 km is, ${\mathrm{s}}_1=100\mathrm{km}/\mathrm{h}$

The speed for the next 40 km is, ${\mathrm{s}}_2=40\mathrm{km}/\mathrm{h}$

The ratio of distance to time is termed speed. It is a scalar quantity.

The expression for the speed is given as,

$\mathrm{s}=\frac{\mathrm{Distance}}{∆\mathrm{t}}$ (i)

Here, s is the speed and $∆t$is the time duration.

The total time for the journey is from 8 am to 11:15 am. That is, $∆\mathrm{t}=3.25\mathrm{h}$

The time taken to travel the first 100 km is,

role="math" localid="1656401191104" $$\begin{gathered} ∆{\mathrm{t}}_1=\frac{\mathrm{Distamce}}{{\mathrm{s}}_1} \\ =\frac{100\mathrm{km}}{100\mathrm{km}/\mathrm{h}} \\ =1\mathrm{h} \end{gathered}$$

The time taken to travel the next 40 km is,

$$\begin{gathered} ∆{\mathrm{t}}_2=\frac{\mathrm{Distamce}}{{\mathrm{s}}_2} \\ =\frac{40\mathrm{km}}{40\mathrm{km}/\mathrm{h}} \\ =1\mathrm{h} \end{gathered}$$

Remaining time for the journey is,

$$\begin{gathered} ∆{\mathrm{t}}_3=∆\mathrm{t}-∆{\mathrm{t}}_1-∆{\mathrm{t}}_2 \\ =3.25\mathrm{h}-1\mathrm{h}-1\mathrm{h} \\ =1.25\mathrm{h} \end{gathered}$$

The remaining distance is,

$$\begin{gathered} \mathrm{Remaining}\mathrm{distance}=300\mathrm{km}-100\mathrm{km}-40\mathrm{km} \\ =160\mathrm{km} \end{gathered}$$

The speed for the remaining trip is calculated as,

$$\begin{gathered} {\mathrm{s}}_3=\frac{\mathrm{Distance}}{∆{\mathrm{t}}_3} \\ =\frac{160\mathrm{km}}{1.25\mathrm{h}} \\ =128\mathrm{km}/\mathrm{h} \end{gathered}$$

Therefore,the least speed needed for the rest of the trip to arrive in time for the interview is128 km/hr .