Posted by nadra on Friday, February 4, 2011 at 1:05pm.
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  nadra, Friday, February 4, 2011 at 3:01pm
a)v(4)=2 seconds
b)a(4).67

calculus  nadra, Friday, February 4, 2011 at 3:02pm
i don't know the answer for partc.

calculus  nadra, Friday, February 4, 2011 at 3:03pm
is answer a and b correct

calculus  Reiny, Friday, February 4, 2011 at 3:39pm
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  MathMate, Friday, February 4, 2011 at 3:43pm
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=4.5t0=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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