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A tugboat is cruising at a speed of 11.4 m/s in still water when it shuts off its engines and coasts to rest. The water exerts drag forces on the tugboat such that its speed during coasting is given by v = vie−ct,where v is the speed at time t, vi is the initial speed at t = 0, and c is a constant. At a time of t = 18.0 s after the engines are cut, the speed is 5.00 m/s. (a) What is the constant c (in s−1) in the expression
(b) What is the speed (in m/s) at t = 40.0 s?
(c) Given the expression above for the speed at any time, which of the expressions below gives the tugboat's acceleration while coasting?

Respuesta :

ogunbasefemi

Answer:

a. c = 0.1739 {m}/{s^2}

b. v(40.0 s ) = 1.64 \frac{m}{s}

Explanation:

So, the equation for the speed is:

v(t) = v_i - c tv(t)=v

i

−ct

as we know that the initial speed is

v_i = 8.2 \frac{m}{s}v

i

=8.2

s

m

and the speed at t=18.4 s is

v(18.4 s ) = 8.2 \frac{m}{s}- c 18.4 s = 5.00 \frac{m}{s}v(18.4s)=8.2

s

m

−c18.4s=5.00

s

m

Now, we can work it a little:

- c 18.4 s = 5.00 \frac{m}{s} - 8.2 \frac{m}{s}−c18.4s=5.00

s

m

−8.2

s

m

- c 18.4 s = -3.20 \frac{m}{s}−c18.4s=−3.20

s

m

- c 18.4 s = -3.20 \frac{m}{s}−c18.4s=−3.20

s

m

c = \frac{ -3.20 \frac{m}{s} }{ - 18.4 s }c =

−18.4s

−3.20

s

m

c = 0.1739 \frac{m}{s^2} c = 0.1739

s

2

m

So, at t=40.0 s the speed will be:

v(40.0 s ) = 8.2 \frac{m}{s}- 0.1739 \frac{m}{s^2} * 40.0 sv(40.0s)=8.2

s

m

−0.1739

s

2

m

∗40.0s

v(40.0 s ) = 8.2 \frac{m}{s}- 6.96 \frac{m}{s}v(40.0s)=8.2

s

m

−6.96

s

m

v(40.0 s ) = 1.64 \frac{m}{s}

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