Question

The children cycled between Camp A and Camp B which are 500 m apart in 50 seconds. If they wanted to reduce the time to 40 seconds, what is the increase in speed?

Answer 1

distance=500 m.
initial time=50 sec.
then speed= dist.$\frac{distance}{time}$
                =$\frac{500}{50}$=10 m/sec.

final time=40 sec.
 then similarly speed = 12.5 m/sec
 change in speed=12.5-10= 2.5 m/sec

Answer 2

speed of the children=d/t=500/50=10m/s.
if t=40s and d=500m
therefore speed=500/40=12.5m/s.
therefore increase in speed = (12.5-10)m/s
=2.5 m/s.