he following is the force as a function of time for an 80 kg object: Determine t
ID: 1778104 • Letter: H
Question
he following is the force as a function of time for an 80 kg object: Determine the impulse on this object from 5 s to 8 s. Determine at what time the force is maximum and what time the force is minimum. At the time when the force is a maximum and when it's a minimum, determine the acceleration vector of the object. The object moves in one dimension along the x-axis. a. b. c. d. If the object has a velocity of 4.05 at 5 s, determine the object's velocity at 8 s (***Be careful, this object is not under constant acceleration!!).Explanation / Answer
F =4t^3 -12t^2+ 8t -1
a) Impulse = integeral ( Fdt) = ( 4t^3 -12t^2+ 8t -1 ) dt
J = ( t^4 - 4t^3 + 4t^2 - t) plugiing the limits,
( 8^4 - 4 x 8^3 + 4 x 8^2 -8 ) - (5^4 - 4 x 5^3 + 4 x 5^2 -5) = ( 2104- 220) = 1884
b) dF/dt = 12t^2 - 24t + 8
12t^2 - 24t + 8 = 0
3t^2 - 6t + 2=
t = 6 +- qroot ( 36 -24)/ 6= ( 6 +- 4.9) / 6
t = (0.183, 1.817 )
F ( (0.183)= 4(0.183)^3 -12x (0.183) ^2+ 8(0.183) -1= 0.00612 - 0.401868 + 0.464= 0.068 N
F (1.817) = 4 ( 1.817)^3 - 12 ( 1.817)^2 + 8 ( 1.817) - 1= - 0.081 N
c) a= F/ m = 0.068 N/ 80 = 8.5x 10^-4 m/s^2 i
a= - 0.081 N / 80 =- 1.01 x 10^-3 m/s^2 i
d) a= F/m =(1/80) (4t^3 -12t^2+ 8t -1)
a= d^2x/dt^2
adt = d ( dx/dt)
dv= adt
V( 8 seonds) - V ( 5 seconds) = integeral ( adt) =( t^4 - 4t^3 + 4t^2 - t )/ 80
V ( 8 seconds ) - 4.05 =23.55
V = 27.6 m/s apprx
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