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Truck problem (Kinematics)

  • #1

Homework Statement



Farhana is late for class and she is ready to do anything to get to school on time. She knows a mattress truck passes her balcony exactly at 8:42 a.m, travelling at a constant 40km/h. If her balcony is 8.21m from the ground and the truck has a height of 3.47. How far must the truck be from the balcony for Farhana to land on it.

Now i know, we have to find Δ d, and v1= 40km/h or 11.1111111111 m/s.

Homework Equations



Δd= v1Δt+ 1/2 aΔt^2


The Attempt at a Solution



Δd= 8.21m- 3.47m
= 4.74m

Δt= Δd/Δv
= 4.74/ 11.111111111
= 0.4266 s

a= -9.8m/s^2

v1= 11.111111111m/s

Now it's asking how far the truck has to be, so it is asking for distance, and distance is speed over time. But i don't know how to find time. i tried but am i doing it right?
 

Answers and Replies

  • #2
28
0
So first step for you is to acknowledge that

time for truck to get to the balcony = time that takes her to fall on top of the truck

now for the girl to fall off can be derived by using the "five kinematic equation"
we know the V1 = 0, we know delta d=4.74 we know acceleration is a=-9.8
we can now solve for the time

now once you have the time
you know the equation distance = speed x time
we know the speed of the truck and now we know time
we can therefore get the distance
 
  • #3
So first step for you is to acknowledge that

time for truck to get to the balcony = time that takes her to fall on top of the truck

now for the girl to fall off can be derived by using the "five kinematic equation"
we know the V1 = 0, we know delta d=4.74 we know acceleration is a=-9.8
we can now solve for the time

now once you have the time
you know the equation distance = speed x time
we know the speed of the truck and now we know time
we can therefore get the distance
I tried doing this and still got the wrong answer :S
 
  • #4
The answer is 11m

and t= 0.569875104 or t= 1.697471834.

if distance = speed * time

and speed is 11.11m/s
dist= 11.11 * 0.569875104
dist= 6.331312405 (6.3m)

and the other one is just wayy off.
 
  • #5
28
0
I got 11 m

did you get time = 0.98 second?

I think you got the quadratic equation wrong

[tex]\Delta d = V_{i}\Delta t + ( 1/2)a(\Delta t)^2 [/tex]

[tex] \Delta t = \frac{-V_{i} \pm \sqrt{ (V_{i})^2 - 4(1/2)a(-\Delta d)}}{a} [/tex]
 
  • #6
did you use v1= 0 or 11.11?
 
  • #7
tell me if i'm doing anything wrong,


[tex]\Delta d = V_{i}\Delta t + ( 1/2)a(\Delta t)^2 [/tex]

delta t = -11.11 +- sqrt (11.11)^2 -4(1/2)(-9.8)(-4.74)
---------------------------------------------
-9.8

-11.11+- sqrt 30.5281
----------------------
-9.8


-11.11+5.525223977
----------------- = 0.569875104
-9.8


-11.11 - 5.525223977
--------------------- = 1.697471834
-9.8

am i right till here?
 
  • #8
28
0
[tex] V_{i} = 0 [/tex]
 
  • #9
28
0
I got 11 m

did you get time = 0.98 second?

I think you got the quadratic equation wrong

[tex]\Delta d = V_{i}\Delta t + ( 1/2)a(\Delta t)^2 [/tex]

[tex] \Delta t = \frac{-V_{i} \pm \sqrt{ (V_{i})^2 - 4(1/2)a(-\Delta d)}}{a} [/tex]
To isolate t to one side



[tex]\Delta d = V_{i}\Delta t + ( 1/2)a(\Delta t)^2 [/tex]


[tex]0 = V_{i}\Delta t + ( 1/2)a(\Delta t)^2 -\Delta d [/tex]

from here you use the quadratic equation which would be

if [tex] y= ax^2 + bx +c [/tex]

[tex] x= \frac{b \pm \sqrt {b^2 -4ac}}{2a} [/tex]

in this case
a= (1/2)(9.8)
b= Vi
c= -( delta d)
x = delta t
 
  • #10
OH OK, thank you so much. I finally now get it!
 
  • #11
28
0
anytime :) I am glad you got it
 

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