Question

A particle was at rest from 9 am to 9:30 am. It moved at a uniform speed of 10 km/hr from 9:30 to 10:00 am. Find the average speed between

1. 9:00 am to 10:00 am
2. 9:15 am to 10:00 am

Answer 1

The explanation for 1:

1. The distance moved by the particle between 9:00 am to 10:00 am is,

$$\begin{gathered} S=vt \\ S=10\times 0.5=5km \end{gathered}$$

2. This is also the distance moved in time 9:00 am to 10:00 am.

3. Thus, the average speed in this interval is,

$averagespeed=\frac{5}{1}=5km/hr$

The explanation for 2:

1. The distance moved between 9:30 am to 10:00 am is 5 km. This is also the distance moved in the interval 9:15 am to 10:00 am. The average speed during this interval is

$V_{avg}=\frac{S}{t}=\frac{5km}{45min}=\frac{5}{{\displaystyle \frac{45}{60}}}=5\times \frac{4}{3}=6.67km/hr$