A person sees in the distance a lightning bolt pass close to a low flying airplane. The person hears thunder 5.6 s after seeing the bolt, and sees the airplane overhead 13 s after hearing the thunder. Find the distance of the airplane from the person at the instant of the bolt. Neglect the time it takes the light to travel from the bolt to the eye. The speed of sound in air is 1100 ft/s. Find the velocity of the airplane

Explanation:

velocity v = Δx/Δt

Δx is the change in displacement

Δt is the change in time

Δx = vΔt

Given v = 1100ft/s

Δt = t2-t1

Δt = 13-5.6

Δt =  7.4s

Δx = 1100*7.4

Δx = 8,140ft

Hence the distance of the airplane from the person at the instant of the bolt is 8,140ft

Since the air moves at constant speed, it means the speed is not chnaging hence the velocity of the air v is the same as the constant speed.

velocity of the airplane vp = Δx/Δt

Δx = 8,140ft

Δt = t2+t1

Δt = 13+5.6

Δt = 18.6s

Substituting the given values into the formula vp = Δx/Δt

vp = 8,140/18.6

vp = 437.63m/s

Hence the velocity of the airplane is 437.63m/s

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