help help!! calc~~~ ^^?

Question

a particle moves along a line so that its postion at any time t is greater than or equal to 0 is given by the function

s(t)=t^2-3t+2,

where s is measured in meters and t is measured in seconds

a) find the displacement during the first 5 seconds

b) find the average velocity during the first 5 seconds

c) find the instantaneous velocity when t =4

d) find the acceleration of the particle when t =4

e) at what values of t does the particle change direction?

f) where is the particle when s is a minimum?

Answer 1

well, you have the position function, s(t).
the velocity function, v(t) is equal to the derivative of the position function.
acceleration function is equal to the derivative of velocity function.
a(t)=v'(t)=s"(t)
so, take first derivative to get velocity function:
s'(t)=v(t)=2t-3
a(t)=v'(t)=2
the particle changes direction at the points where the particle stops.
you can find this by figuring out the zeros of the v(t) function. then, see if those same values of t makes the a(t) function go to zero. it is only changing direction where v(t)=0 and a(t) does not =0.
to find the instantaneous velocity, plug in t=4 into the velocity function (you can do the calculations out urself, its easy).
average velocity would have to be calculating the value of the function in the time interval [0,5] and dividing by 2.
goodluck.

Answer 2

a) s(t=5) = 5^2 -3(5) + 2 = 25 - 15 + 2 = 12
b) velocity ==v= d/dt[s(t)] = 2t - 3
v(1) = 2(1)-3 = -1
v(2) = 2(2)-3 = 1
v(3) = 2(3)-3= 3
v(4) = 2(4)-3 = 5
v(5) = 2(5)-3 = 7
Average velocity = (-1+1+3+5+7)/5 = 15/5 = 3
c) v(4) = 5 see above
d) d^2/dt^2[s(t)] = 2 = accleration for all t
e) between 1 and -1
f) s(t) is a minimum when the first derivative, 2t-3=0 or t=2/3
s(t=2/3) = (2/3)^2 -3(2/3) + 2 = 4/9 - 2 + 2 = 4/9

Answer 3

A) just plug in t=1, t=2, t=3 etc. so on and so forth to get the displacement.
B) Find the derivative of the equation u have and then plug in t=1,t=2, etc, into the derivative equation u just got.
C)plug in t=4 into the derivative of the s(t) functions [by the way u found this in part b the equation]
D)find the derivative of the VELOCITY FUNCTION then plug in t=4
FYI
IF U R GIVEN A S(T) OR BASICALLY DISTANCE TIME FUNCTION THE DERIVATIVE OF THAT FUNCTION GIVES U A VELOCITY FINCTION AND THE DERIVITIVE OF THE VELOCITY FUNCTION GIVES U THE ACCELERATION FUNCTION.
e) for this ur going to have to make a interval chart(if u havent been taught then i really cant teach it should be in ur book) but like do the chart if u dont get this part i guess email me IF i can i will make one to show u how to do it. but like u put in the velocity and the acceleration functions in and if u get a MINUS ( - ) then it means it's direction changed.
f) i think for this one u plug in like 0 and then 1 whichever one is smaller is the minimum .
[e-mail]

Answer 4

Why is this so hard?
You have the position equation.
Let's see if I can remember.
Velocity is the first derivative:
2t-3
And acceleration is constant:
2

Answer 5

surely in your equation,you left out,r.