Asked by laura
What is the acceleration of the particle the first time its velocity equals zero?
s(t)= -4sin(t) - (t/2) +10
What is the acceleration of the particle the first time its velocity equals zero?
−5.197
0.745
1.323
2.550
3.969
Could somebody write out the steps and explain this to me?
s(t)= -4sin(t) - (t/2) +10
What is the acceleration of the particle the first time its velocity equals zero?
−5.197
0.745
1.323
2.550
3.969
Could somebody write out the steps and explain this to me?
Answers
Answered by
Damon
have to know how many time derivatives to take.
Answered by
laura
A particle moves along a line so that at time t, where 0 < t < π, its position is given by
Sorry! this was the first part
Sorry! this was the first part
Answered by
Damon
s(t)= -4sin(t) - (t/2) +10
v = ds/dt = -4cos t -1/2
a = d^2s/dt^2 = 4 sin t
what is t the first time v = 0?
4 cos t = -1/2
cos t = -1/8
t = 97.2 deg or 360 -97.2
it says the first time so
t = 97.2 deg * pi/180 = 1.7 radians
so what is a?
4 sin 97.2 = 3.97 looks like the last one
v = ds/dt = -4cos t -1/2
a = d^2s/dt^2 = 4 sin t
what is t the first time v = 0?
4 cos t = -1/2
cos t = -1/8
t = 97.2 deg or 360 -97.2
it says the first time so
t = 97.2 deg * pi/180 = 1.7 radians
so what is a?
4 sin 97.2 = 3.97 looks like the last one