Question

h(t) 0.6cos(2t) + 08 sin(t)
h'(t) = v(t) = -1.2sin(2t) + 0.8cos(t)

find the vertical displacement when the velocity is 0.8m/s

how do i solve v(t) = 0.8m/s

I tried the normal way by equating and factoring, however two answers are obtained and the second method is the identity: asin(x) + bcos(x) = Rsin(x+theeta). In the identity to find theeta should i consider the -1.2 or just its absolute value? because then i get t=0 which is the only answer.

Answer 1

Perhaps there is an easier way, but this is what I suggest for now:

h'(t) = v(t) = -1.2sin(2t) + 0.8cos(t) =0.8

-1.2sin(2t) + 0.8cos(t) = 0.8
Divide by 0.8 throughout to get
-1.5sin(2t) + cos(t) = 1

expand sin(2t) by the double angle formula and factor out cos(t)
-3sin(t)cos(t) + cos(t) = 1
cos(t)(1-3sin(t)) = 1

Substitute c=cos(t) and √(1-c²)=sin(t):
c(1-3√(1-c²)) = 1

square (and verify roots later)
(c-1)²=9c²(1-c²)

We know that c=1 is a factor, so do a long division and solve the resulting cubic to get
c1=1, c2=(sqrt(19)/3^(9/2)+1/27)^(1/3)+2/(27*(sqrt(19)/3^(9/2)+1/27)^(1/3))-1/3=0.25643078474621
or you can solve for c2 numerically.
The other two roots are complex, so they are rejected automatically.

Now solve for t:
t1=acos(c1)=acos(1) = 0
t2=±acos(c2)=±acos(0.25643078474621)
=±1.311468632085831
substitute into v(t):
v(0)=0.8
v(1.311468632085831)=-0.3897, rejected.
v(-1.311468632085831) = 0.8 OK
To get a positive value for t2, add 2π to get
t2=4.971716675093755