Question

A car traveller at an uniform speed. At 2 pm it is at a distance of180 on and at 6 pm it is at 360 km. Using section formula find at what distance it will reach 12 mid night.

Answer 1

Answer: 630 km

1st time, t₁ = 2 pm

2nd time, t₂= 6 pm

∴ Δt= (t₂-t₁)= 4 hours, that means the difference between 1st and 2nd time is 4 hours.

Now,

d₁= 180 km

d₂= 360 km

∴ Δd =(d₂-d₁)=(360-180)=180 km

So,

velocity, v= Δd÷Δt= (180÷4) km/h (kilometer per hour) = 45 km/h

3rd time, t₃= 12 am

difference between 2nd and 3rd time Δt' = t₃-t₂= 12 am - 6 pm =  6 hours

Let, total distance= d₃

Now, we got

velocity, v = 45 km/h

time, Δt' = 6 h

distance Δd' =?

we know,

v = Δd'÷Δt'

or, Δd' = v × Δt

∴ Δd' = 45 × 6 =270 km

∴ Total distance d₃= Δd'+d₂= (270+360) = 630 km