a particle moves along a number line measured in cm so that its position at time t sec is given by s=72/(t+2) +k, k is a constant and t>=0 seconds.

(a) Find the instantaneous velocity of the particle at t=4 seconds
(b) Find the acceleration of the particle when t =4 seconds
(c) If we know the particle is at s=20 when t=4sec, use your answer from part( a) to approximate the position of the particle at t=4.5 sec.

a)v(4)=-2 seconds

b)a(4).67

i don't know the answer for partc.

is answer a and b correct

a) and b) are correct

for c)
sub in the given values
20 = 72/6 + k
20 = 12 + k
k = 8

so now you know
s= 72/(t+2) + 8
when t = 4/5
s = 72/4.5 + 8 = 24

Part (a) is correct, and (b) is correct to 2 decimal places.

For part (c), you can make use of the approximation formula
f(t0+h)=f(t0)+h*f'(t0) approximately, and when h is relatively small.
t0=4
f(t0)=20,
f'(t0)=-2
h=4.5-t0=0.5
f(t0+h)=f(4.5)=20+(-2)*0.5=19 (approx.)

Check by initial conditions:
f(4)=20
substitute in s(t) to get
20=72/(4+2)+k=12+k
=> k=8
So
f(t)=72/(t+2)+8
f(4.5)=72/(4.5+2)+8
=19.077...
So approximation above is reasonably accurate.

To find the instantaneous velocity of the particle at t=4 seconds, we need to find the derivative of the position function with respect to time.

(a) Find the derivative of the position function:

s(t) = 72/(t+2) + k

Using the quotient rule, the derivative of s(t) with respect to t is:

s'(t) = (-72/(t+2)^2) * (1) = -72/(t+2)^2

Now, substitute t=4 into the derivative to find the instantaneous velocity at t=4 seconds:

s'(4) = -72/(4+2)^2 = -72/36 = -2 cm/s

So, the instantaneous velocity of the particle at t=4 seconds is -2 cm/s.

(b) To find the acceleration of the particle at t=4 seconds, we need to find the second derivative of the position function:

s''(t) = d^2s/dt^2 = d/dt (-72/(t+2)^2)

Using the chain rule and the power rule for differentiation, we have:

s''(t) = (-2) * (-72) * (d/dt) * (1/(t+2)^3)
= 144/(t+2)^3

Substitute t=4 into the second derivative to find the acceleration at t=4 seconds:

s''(4) = 144/(4+2)^3 = 144/216 = 2/3 cm/s^2

So, the acceleration of the particle at t=4 seconds is 2/3 cm/s^2.

(c) To approximate the position of the particle at t=4.5 seconds using the instantaneous velocity at t=4 seconds, we can use the concept of average velocity over a small time interval.

The average velocity over a small time interval is given by the change in position divided by the change in time:

Average velocity = (s(4.5) - s(4))/(4.5 - 4)

We can approximate the change in position s(4.5) - s(4) using the instantaneous velocity at t=4 seconds:

Approximate change in position = Instantaneous velocity at t=4 seconds * (4.5 - 4)

Substituting the values:

Approximate change in position = (-2 cm/s) * (4.5 - 4) = -2 cm

We are given that the particle is at s=20 when t=4 seconds. So, to find the approximate position at t=4.5 seconds, we can add the approximate change in position to the position at t=4 seconds:

Approximate position at t=4.5 seconds = s(4) + Approximate change in position
= 20 cm + (-2 cm) = 18 cm

Therefore, the approximate position of the particle at t=4.5 seconds is 18 cm.