Question

A train usually covers a journey of 240 km at a steady speed of v km h -1 . One day, due to adverse weather condition , it reduces its speed by 40 kmh-1. And jouney takes one hour longer. Derive v2-40v-9600=0

Answer 1

d = 240 km
Tn=Tnew=Told+1, where Told is the time it originally took to go 240 km
d = V*Told=Vnew*Tn, where V=original velocity, Vnew is new velocity (i.e. V-40)
V*Told=(V-40)*(Told+1), since distance traveled going slow and fast is the same (i.e. 240 km)

=V*Told + V - 40Told -40
V(Told-Told-1)=-40 - 40Told
-V=-40(1+Told)
V=40(1+Told)

d=V*Told=40(1+Told)Told
=40Told^2+40Told=240
=Told^2 + Told - 6 = 0
=(Told +3)(Told -2)=0

So, Told=-3 or Told = 2

This implies that V = d/Told = 240/-3 = -80 or V=240/2 = 120.

This solution can be represented algebraically as (V-120)(V+80)=0
(V-120)(V+80)=0
=V^2+80V-120V-9600=0
=V^2-40V-9600=0