Answer:
The pickup truck and hatchback will meet again at 440.896 m
Explanation:
Let us assume that both vehicles are at origin at the start means initial position is zero i.e. 
= 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.
Truck:
= 33.2 m/s, a = 0 (since the velocity is constant), = 0
Using 
s = 33.2t .......... eq (1)
Hatchback:
, = 0 m/s (since initial velocity is zero), = 0
Using
putting in the data we will get
now putting 's' value from eq (1)
which will give,
t = 13.28 s
so both vehicles will meet up gain after 13.28 sec.
putting t = 13.28 in eq (1) will give
s = 440.896 m
So, both vehicles will meet up again at 440.896 m.
Answer:
use the kinematics equations and solve for time, after that use dleta dx=Vi*delta time
Before swinging, T has only potential energy, (no speed)
Ui = mgh
Where h is the vertical displacement of T
From the laws of geometry,
cos45 = (L-h)/L
cos45 = 1-h/L
h/L = 1-cos45
h = L(1-cos45)
Therefore
Ui = mgL(1-cos45)
Proceeding the same way,
Twill raise to aheight of h' due to swing
h' = L(1-cos30)
The PE of T after swing is
Uf = mgh'
Uf = mgL(1-cos30)
Along with the PE , T has some kinetic energy results due to the moment.
Tf = 0.5*mv^2
According to the law of conservation of energy,
Ui = Uf+Tf
mgL(1-cos45) = mgL(1-cos30) + 0.5*mv^2
gL(co30-cos45) = 0.5*v^2
9.8*20*(co30-cos45) = 0.5*V^2
v = 7.89 m/s
<span>The speed f T after swing is 7.89 m/s</span>
