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=4.5-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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