Solve Kinematics Problem: Stone Thrown from 165m Building, 4s Travel Time

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SUMMARY

The discussion centers on solving a kinematics problem involving a stone thrown downward from a 165 m building with an initial velocity of 5.0 m/s. The correct approach to find the displacement during the 4th second of travel requires understanding that it is not the total distance after 4 seconds but rather the distance covered specifically in that second. The correct displacement during the 4th second is calculated using the formula for displacement in a given time interval, leading to a final answer of 39 m, as clarified by participants in the discussion.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = vit + 1/2at²
  • Knowledge of initial velocity and acceleration due to gravity (9.81 m/s²)
  • Ability to differentiate between total distance and displacement during specific time intervals
  • Familiarity with basic physics concepts related to motion
NEXT STEPS
  • Study the concept of displacement versus distance in kinematics
  • Learn how to apply kinematic equations to solve for displacement in specific time intervals
  • Explore the implications of initial velocity and acceleration in vertical motion problems
  • Practice additional problems involving free fall and projectile motion
USEFUL FOR

Students studying physics, particularly those focusing on kinematics, as well as educators seeking to clarify concepts related to motion and displacement.

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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.
 
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Well, d=volt+.5at^2
so d should be 98.4m. Did you get that?
 
The answer given is 39 m.
 
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
 
Last edited:
d = displacement/distance.
 
Yes, i know.
 
Is anyone going to help solve this problem please?
 
I just gave you the answer.
 
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?
 

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