Stone thrown downward from 165m building: distance in 4th second

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Homework Statement

A student throws a stone vertically downward with an initial velocity of 5.0 m/s from the top of a 165 m building. How far does the stone travel during its 4th second of travel?

Homework Equations

d= vit + 1/2at2

The Attempt at a Solution

Vi = -5.0 m/s
t = 4.0 s
a = -9.81 m/s2
d = ?

I tried it in the kinematics equation d= vit + 1/2at2 and the answer is wrong.

 

[silvashadow]

Well, d=volt+.5at^2
so d should be 98.4m. Did you get that?

 

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The answer given is 39 m.

 

[silvashadow]

Wait, I think it was the d for the 4th second, from 3-4s. Because the first second is 0-1s, second second is for 1-2s, etc.

So
Vf=vo+at
V@3=5+9.8*3 and V@4=5+9.8*4
Vf^2=Vo^2+2ad
(V@4)^2=(V@3)^2+2*9.8*d
Find d

 

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d = displacement/distance.

 

[silvashadow]

Yes, i know.

 

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Is anyone going to help solve this problem please?

 

[silvashadow]

I just gave you the answer.

 

[Saladsamurai]

As silvashadow has already pointed out, you are not reading the question carefully. It does not ask for the distance after 4 seconds, it asks for the displacement during its 4th second

Displacement during any given time interval=
(Position at final time)-(Position at initial time).

What is the final time? What is the initial?