Exercício de Dinamica (103)

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114 
 
PROBLEM 11.80 
During a manufacturing process, a conveyor belt starts from rest and travels a total of 1.2 ft before 
temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to ±4.8 ft/s2 per 
second, determine (a) the shortest time required for the belt to move 1.2 ft, (b) the maximum and average 
values of the velocity of the belt during that time. 
 
SOLUTION 
Given: At max0, 0, 0; 1.2 ft;t x v x= = = = 
 when 2
max
max
, 0; 4.8 ft/s
da
x x v
dt
 = = = 
 
 
(a) Observing that maxv must occur at 1
min2 ,t t= the a t− curve must have the shape shown. Note that the 
 magnitude of the slope of each portion of the curve is 4.8 ft/s2/s. 
 
 We have at :t t= Δ max max
1 1
0 ( )( )
2 2
v t a a t= + Δ = Δ 
 2 :t t= Δ max max max max
1 1
( )( )
2 2
v a t t a a t= Δ + Δ = Δ 
 Using symmetry, the v t− is then drawn as shown. 
 
 Noting that 1 2 3 4A A A A= = = and that the area under the v t− curve is equal to max,x we have 
 max max
2
max max max max
(2 )( )
2
t v x
v a t a t x
Δ =
= Δ  Δ =