Answer :
Answer:
time = 182.68 s
Explanation:
given data
truck approaches = 470 Hz
truck recedes = 400 Hz
position away = 5.0 km
solution
if V is the speed of sound in air and v is the speed to truck
f is the frequency of sound
when truck approach
f1 = [tex]\frac{V+v}{V} f[/tex]
and when truck recede
f2 = [tex]\frac{V+v}{V} f[/tex]
so here we can say
[tex]\frac{f1}{f2} = \frac{V+v}{V-v}[/tex] ...........1
[tex]\frac{470}{400} = \frac{340.278+v}{340.278-v}[/tex]
v = 27.37 m/s
so here 5 km
S = u×t + 0.5 ×a×t² ...........2
5 × 10³ = 27.37 t + 0
t = 182.68 s