A train is moving parallel and adjacent to a highway with a constant speed of 25 m/s. A car, at time t0, is 42 m behind the train, traveling in the same direction as the train at 49 m/s, and accelerating at 4 m/s^2. The train's whistle blows at a frequency of 460Hz. The speed of sound in air is 343 m/s. At the moment t0, the car's driver hears a whistle from the train.
What frequency does the car's driver hear?
Part B. @ a time t, after the car has passed the train and is a distance 32 m ahead of the train, the whistle blows again. What frequency does the driver hear?
f' =f [ (1+ v(observer) / v) / (1- v(source) / v)] <---equation 1
x(final) = x(initial) + v t +.5 a t2 <---equation 2
f'= f [1- (v(observer)/v)] <---equation 3
The Attempt at a Solution
460[(1+(49/343)) / (1-(25/343)) = 567.044 = WRONG ANSWER
I treated the train as stationary to find the time,I determined the car to be traveling 24 m/s
So using equation 2
74 = 0 + 24 t + ( .5 * 4 ) t2
v(final)=49 + ( 2.54 * 4 ) = 59.16 m/s
then using equation 3 I did
f'=460(1- (59.16/343) ) = 380.66=WRONG