Question

An athlete's usual running speed in still air is 8 m/s.
On one windy day, the time he took to run 100m, running with the wind, was 6 2/3 seconds less than the time he took to run 100m, running against the wind.

a. Given that the speed of the wind that day was "w m/s", write down an expression in terms of w for the time he took to run 100m when running with the wind.

b. Form an equation in w and show that it reduces to w^2+30w-64=0

c. Solve the equation to find the speed of the wind

Answer 1

i say b..

Answer 2

Ummm... It's not a multiple choice question, those are actually the questions, you know, a,b and c

Answer 3

(time taken to travel against wind) - (time take to travel with wind) = 6 2/3 seconds

Answer 4

time = distance/speed

Answer 5

can you calculate both the times, and plug in ?

Answer 6

Speed of the man running with the wind = (8+w)
Speed when running against the wind = (8-w)

Time taken for 100m running with the wind = 100/(8+w)
Time taken for 100m running against the wind = 100/(8-w)

Given:
\[\frac{100}{8+w} = \frac{100}{8-w} - 6\frac{2}{3}\]

Can you proceed further ?

Answer 7

the equation can be deduced from ((100/(8-w)) - ((100/(8+w)) =20/3
which reduces to 2w/((8-w)(8+w)) =1/15 and hence w^2+30w-64=0
solving we have (w+32)(w-2)=0
w=2 (w=-32 ignored) spped of wind =2m/s