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Aliun [14]
2 years ago
14

A pontoon boat travels across a lake in a straight line and increases in speed uniformly from vi = 18.5 m/s to vf = 36.0 m/s in

a displacement Δx of 250 m. We wish to find the time interval required for the boat to move through this displacement. g
(a) Draw a coordinate system for this situation.

(b) What

analysis model is most appropriate for describing this situation?

(c) From the analysis model, what equation is most

appropriate for finding the acceleration of the speedboat?

(d) Solve the equation selected in part (c) symbolically for

the boat’s acceleration in terms of vi , vf , and ?x. (e) Substitute

numerical values to obtain the acceleration numerically.

(f) Find the time interval mentioned above.
Physics
1 answer:
omeli [17]2 years ago
8 0

Answer:

Explanation:

a ) The motion is one dimensional , so motion is along x - axis , starting from origin ( 0 , 0 )

b ) Initial velocity is 18.5 m /s when boat is situated at origin . When he displaces by 250 m along x axis and his position is ( 250 , 0 ) along x axis , his velocity becomes 36 m /s . Both his velocity and acceleration is along x - axis.

c ) Initial velocity vi = 18.5 m /s

final velocity vf = 36 m/s

Displacement x  = 250 m

Acceleration a = ?

Most appropriate formula is given below .

vf² = vi² + 2 a x

2ax = vf² - vi²

x =  ( vf² - vi² ) / 2 a

d )

Putting the given values

36² - 18.5² / 2 x 250

= 1296 - 342.25 / 500

= 1.9 m /s².

f ) Time interval t = ?

Required formula

vf = vi + at

t = (vf - vi ) / a

Putting the values

t = (30 - 18.5) / 1.9

= 6.05 second .

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makvit [3.9K]

Answers:

(a) 2509.98 m/s

(b) 397042.215 m

(c) 1917.76 m/s

Explanation:

The question is incomplete, please remember to write the whole question :) However, part (a) is written below:

(a) What is the escape speed on a spherical asteroid whose radius is 700 km  and whose gravitational acceleration at the surface is a_{g}=4.5 m/s^{2}

Knowing this, let's begin:

a) In this part we need to find the escape speed V_{e} on the asteroid:

V_{e}=\sqrt{\frac{2GM}{R}} (1)

Where:

G is the universal gravitational constant

M is the mass of the asteroid

R=700 km=700(10)^{3} m is the radius of the asteroid

On the other hand we know the gravitational acceleration is a_{g}=4.5 m/s^{2}, which is given by:

a_{g}=\frac{GM}{R^{2}} (2)

Isolating GM:

GM=a_{g}R^{2} (3)

Substituting (3) in (1):

V_{e}=\sqrt{\frac{2a_{g}R^{2}}{R}}=\sqrt{2a_{g}R} (4)

V_{e}=\sqrt{2(4.5 m/s^{2})(700(10)^{3} m)} (5)

V_{e}=2509.98 m/s (6) This is the escape velocity

b) In this part we will use the Conservation of mechanical energy principle:

E_{o}=E_{f} (7)

Being:

E_{o}=K_{o}+U_{o}=\frac{1}{2}m V^{2} - \frac{GMm}{R} (8)

E_{f}=K_{f}+U_{f}=0 - \frac{GMm}{R+h} (9)

Where:

E_{o} is the initial mechanical energy

E_{f} is the final mechanical energy

K_{o} is the initial kinetic energy

K_{f}=0 is the final kinetic energy

U_{o} is the initial gravitational potential energy

U_{f} is the final gravitational potential energy

m is the mass of the object

V=1510 m/s is the radial speed of the object

h is the distance above the surface of the object

Then:

\frac{1}{2}m V^{2} - \frac{GMm}{R}=- \frac{GMm}{R+h} (10)

Isolating h:

h=\frac{2 a_{g} R^{2}}{2a_{g}R-V^{2}}-R (11)

h=\frac{2 (4.5 m/s^{2}) (700(10)^{3} m)^{2}}{2(4.5 m/s^{2})(700(10)^{3} m)-(1510 m/s)^{2}}-700(10)^{3} m (11)

h=397042.215 m (12) This is the distance above the asteroid's surface

c) We will use the Conservation of mechanical energy principle again, but now the condition is that the object is dropped at a distance h=981.8 km=981.8(10)^{3} m. This means that at the begining the object only has gravitational potential energy and then it has kinetic energy and gravitational potential energy:

\frac{-GMm}{R+h}=\frac{-GMm}{R}+\frac{1}{2}mV^{2} (13)

Isolating V:

V=\sqrt{2a_{g} R(1-\frac{R}{R+h})} (14)

V=\sqrt{2(4.5 m/s^{2}) (700(10)^{3} m)(1-\frac{700(10)^{3} m}{700(10)^{3} m+981.8(10)^{3} m})} (15)

Finally:

V=1917.76 m/s

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Explanation:

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