What is the speed of the car given the time taken to receive an echo? I am trying to solve this question-

The driver of an engine produced a whistle sound from a distance $800m$ away a hill to which the engine was approaching.The driver heard the echo after $4.5s$.Find the speed of the car is speed of sound through air is $340 m/s$.

My attempt-

$V=\frac {2d}{t}$ (2d since echo sound has to travel twice from A to B and then from B to A again.)
Putting $v=340m/s,d=800m$ we get,$t=\frac{80}{17}s$.
So,time taken to move $2AB$ (AB+AB)=$\frac{80}{17}s$.
Let,driver has moved to point O when sound reaches him at O.
So,time taken by sound to reach O (A to B and then from B to O)=4.5s.
So,2AB time-(AB+OB) time=$(\frac{80}{17})-4.5=\frac{7}{34}s$
So,AB-OB time=$\frac{7}{34}s.$
So,$\frac{40}{17}-OB=\frac{7}{34}s.$
Solving we get,time taken by sound to reach $OB=\frac{73}{34}s$. 
So,distance $OB=speed\times time=340\times \frac{73}{34} m.$
So,we can find AO and OB.
So,time taken by sound to reach from A to O by car=time taken by sound to reach from B to O.
applying time=distance/speed,and making the time taken by sound=that of car we get,
$\frac{AO}{Speed_{car}}=\frac{OB}{Speed_{sound}}.$
Solving we get,$Speed_car=\frac{2380}{34} m/s=32 m/s$ approx.
But answer given is $15.5m/s$.Where am I going wrong?
Thanks for any help.
 A: See sound will travel $340*4.5=1530m$ but in this time train will also travel some distance. original distance between hill and train back and forth is $1600m$ but sound was heard at $1530m$ from hill ie $70m$ from original place of train so train travelled $70m$ in $4.5s$ thus speed is approximately $70/4.5=15.55m/s$ 
A: During these $\Delta t = 4.5$ seconds:


*

*The sound travelled $d_0$ to the hill and $d_1$ back to the driver

*The driver travelled some distance $d_2$ towards the hill


We know that $d_0 = 800$, and that $d_2 = v_{car}*\Delta t$ because the driver travels at constant speed.
Moreover, we know that $d_0 + d_1 = v_{sound}*\Delta t$ because sound travels at constant speed too.
Then, we know that $d_0 - d_2 = d_1$: the sound is heard when it is at the same position as the driver.
Solving for $v_{car}$ gives us:
$v_{car} = \frac{d_2}{\Delta t} = \frac{d_0 - d_1}{\Delta t} = \frac{2 d_0 - v_{sound}\Delta t}{ \Delta t}$
Numerical application gives the desired 15.5 meters per second.
