Q11P - Fundamentals Of Physics

Question

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?

Step-by-Step Solution

Answer

The speed needed for the rest of the trip is 128 km/h .

The least speed needed for the rest of the trip to arriving in time is 128 km/h

Step 1: Given Data

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, ${\mathrm{s}}_2=40 \mathrm{km}/\mathrm{h}$ 

Step 2: Understanding the speed in terms of distance and time.

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.

Step 3: Determination of the remaining time and distance

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, 

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

 

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,

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

Step 4: Determination of the least speed required for the rest of the trip

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 is 128 km/hr .