You are to drive 300km to an interview. The interview is at 11.15 AM. You plan to drive at 100 km/h , so you leave at 8 AM, to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 40 km/h for 40km. What would be the least speed needed for the rest of the trip to arrive in time for the interview?

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 .