Asked by Rachel
A particle moves along a line so that at time t, where 0<=t<=pi, its position is given by
s(t)= -4cost-((t^2)/2)+10 What is the velocity of the particle when its acceleration is zero?
a. 8.13
b. 1.32
c. 0.74
d. -5.19
e. 2.55
I know that c and b are incorrect
Please help me
s(t)= -4cost-((t^2)/2)+10 What is the velocity of the particle when its acceleration is zero?
a. 8.13
b. 1.32
c. 0.74
d. -5.19
e. 2.55
I know that c and b are incorrect
Please help me
Answers
Answered by
bobpursley
Your function is odd, I read it as
Cosine ( t-((t^2/2) ) + 10
check that
Take the derivative, that is velocity
Take the derivative again, and set to zero, solve for t. Put that t into velocity .
Cosine ( t-((t^2/2) ) + 10
check that
Take the derivative, that is velocity
Take the derivative again, and set to zero, solve for t. Put that t into velocity .
Answered by
Fernando
Solve for v(t)
V(t) =4sin(t)-t
Solve for a(t)
a(t)=4cos(t)-1
Find when a(t)=0
t=1.32
Plug t into v(t)
ANSWER=2.55
V(t) =4sin(t)-t
Solve for a(t)
a(t)=4cos(t)-1
Find when a(t)=0
t=1.32
Plug t into v(t)
ANSWER=2.55
Answered by
nom
T'is actually 3.969. Thank me later
Answered by
anon
3.969
Answered by
rachel m
it is 2.55
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.