A student normally needs 15 seconds for a 100 m sprint One d

A student normally needs 15 seconds for a 100 m sprint
One day favorable wind conditions help her to increase her speed by 2 km/h
What is her time now?

st=d
15s=100
s=100/15
=20/3 m/sec
.
2km/h
=2000/60^2
=2000/3600
=20/36
=5/9 m/sec
.
20/3 + 5/9
=60/9 + 5/9
=65/9 m/sec new speed.
.
65t/9=100
t=900/65=180/13 sec
=13.85 sec