# calculus

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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.

• calculus -

a)v(4)=-2 seconds
b)a(4).67

• calculus -

i don't know the answer for partc.

• calculus -

is answer a and b correct

• calculus -

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

• calculus -

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.

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