Question

1. in 1986, the first flight around the globe without a single refueling was completed. the aircraft’s average speed was 186 km/hr. if the airplane landed at this speed and accelerated at -15 m/s², how long did it take for the airplane to stop?
2. the first permanent public railway was built by george stephenson and opened in cleveland, ohio, in 1825. the average speed of the trains was 24.0 km/hr. suppose a train moving at this speed accelerates at -0.20 m/s² until it reaches a speed of 8.0 km/hr. how long does it take the train to undergo this change in speed?
3. the winding cages in mine shafts are used to move workers in and out of the mines. these cages move much faster than any of the other commercial elevators. in one south african mine, speeds of up to 65.0 km/hr are attained. the mine has a depth of 2072 m. suppose two cages start their downward journey at the same moment. the first cage immediately attains the maximum speed (an unrealistic situation), then proceeds to descend uniformly at that speed all the way to the bottom. the second cage starts at rest and then increases its speed with a constant acceleration of magnitude 4.00 x 10⁻² m/s². how long will the trip take for each cage? which cage will reach the bottom of the mine shaft first?
4. mary rife, of texas, set a women’s world record for sailing. in 1977, her vessel, proud mary, reached a speed of 317 km/hr. suppose it takes 8.0 s for the boat to decelerate from 317 km/hr to 200 km/hr. what is the boat’s acceleration? what is the displacement of the proud mary as it slows down?
5. in 1994, a human-powered submarine was designed in boca raton, florida. it achieved a maximum speed of 3.06 m/s. suppose the submarine starts from rest and accelerates at 0.800 m/s² until it reaches maximum speed. the submarine then travels at constant speed for another 5.00 s. calculate the total distance traveled by the submarine.
6. the highest speed achieved by a standard nonracing sports car is 350 km/hr. assuming that the car accelerates at 4.00 m/s², how long would this car take to reach its maximum speed if it is initially at rest? what distance would the car travel during this time?
7. the world’s fastest warship belongs to the united states navy. this vessel, which floats on a cushion of air, can move as fast as 170 km/hr. suppose that during a training exercise the ship accelerates at 2.67 m/s², so that after 15.0 s its displacement is 600 m. calculate the ship’s initial velocity. assume that the ship moves in a straight line.
8. the first supersonic flight was performed by then capt. charles yeager in 1947. he flew at a speed of 300 m/s at an altitude of more than 12 km, where the speed of sound in air is slightly less than 300 m/s. suppose capt. yeager accelerated 7.20 m/s² in 25.0 s to reach a final speed of 300 m/s. what was his initial speed?
9. in 1976, kitty hambleton of the united states drove a rocket-engine car to a maximum speed of 965 km/hr. suppose kitty started at rest and underwent a constant acceleration with a magnitude of 4.0 m/s². what distance would she have to travel in order to reach the maximum speed?
10. with a cruising speed of 2.30 x 10³ km/hr, the french supersonic passenger jet concorde is the fastest commercial airplane. suppose the landing speed of the concorde is 20.0% of the cruising speed. if the plane accelerates at -5.80 m/s², how far does it travel between the time it lands and the time it comes to a complete stop?

Answer:

Let's solve problem 1 step by step:

Problem 1:

Given:

• Initial speed, \( u = 186 \, \text{km/hr} \) (convert to \( \text{m/s} \): \( 186 \times \frac{1000}{3600} = 51.67 \, \text{m/s} \))
• Acceleration, \( a = -15 \, \text{m/s}^2 \) (negative because it’s deceleration)
• Final speed, \( v = 0 \, \text{m/s} \) (stops)

We use the kinematic equation:
\[ v = u + at \]

Step 1: Rearrange for time \( t \)

\[ t = \frac{v - u}{a} \]

Step 2: Substitute values

\[ t = \frac{0 - 51.67}{-15} \]
\[ t = \frac{-51.67}{-15} \approx 3.44 \, \text{seconds} \]

Final Answer (Problem 1):

The airplane takes approximately \( \boldsymbol{3.44 \, \text{s}} \) to stop.

Problem 2:

Given:

• Initial speed, \( u = 24 \, \text{km/hr} \) (convert to \( \text{m/s} \): \( 24 \times \frac{1000}{3600} \approx 6.67 \, \text{m/s} \))
• Acceleration, \( a = -0.20 \, \text{m/s}^2 \) (deceleration)
• Final speed, \( v = 8.0 \, \text{km/hr} \) (convert to \( \text{m/s} \): \( 8.0 \times \frac{1000}{3600} \approx 2.22 \, \text{m/s} \))

Using \( v = u + at \), solve for \( t \):

Step 1: Rearrange

\[ t = \frac{v - u}{a} \]

Step 2: Substitute

\[ t = \frac{2.22 - 6.67}{-0.20} \]
\[ t = \frac{-4.45}{-0.20} \approx 22.25 \, \text{seconds} \]

Final Answer (Problem 2):

The train takes approximately \( \boldsymbol{22.3 \, \text{s}} \) (rounded) to slow down.

Problem 3:

Cage 1 (constant speed):

• Depth, \( d = 2072 \, \text{m} \)
• Speed, \( v = 65.0 \, \text{km/hr} = 65.0 \times \frac{1000}{3600} \approx 18.06 \, \text{m/s} \)

Time for Cage 1:
\[ t_1 = \frac{d}{v} = \frac{2072}{18.06} \approx 114.7 \, \text{s} \]

Cage 2 (accelerated then constant speed):

• Acceleration, \( a = 4.00 \times 10^2 \, \text{m/s}^2 \) (wait, this is unrealistic—probably a typo, assume \( 4.00 \, \text{m/s}^2 \))
• Let’s assume it accelerates to maximum speed (same as Cage 1: \( 18.06 \, \text{m/s} \)) first.

Time to accelerate (\( t_{\text{accel}} \)):
\[ v = u + at_{\text{accel}} \implies t_{\text{accel}} = \frac{v - u}{a} = \frac{18.06 - 0}{4.00} \approx 4.515 \, \text{s} \]

Distance during acceleration (\( d_{\text{accel}} \)):
\[ d_{\text{accel}} = ut_{\text{accel}} + \frac{1}{2}at_{\text{accel}}^2 = 0 + 0.5 \times 4.00 \times (4.515)^2 \approx 40.8 \, \text{m} \]

Remaining distance (\( d_{\text{remain}} \)):
\[ d_{\text{remain}} = 2072 - 40.8 = 2031.2 \, \text{m} \]

Time to travel remaining distance (\( t_{\text{remain}} \)):
\[ t_{\text{remain}} = \frac{d_{\text{remain}}}{v} = \frac{2031.2}{18.06} \approx 112.5 \, \text{s} \]

Total time for Cage 2:
\[ t_2 = t_{\text{accel}} + t_{\text{remain}} \approx 4.515 + 112.5 \approx 117.0 \, \text{s} \]

Final Answer (Problem 3):

• Cage 1 time: \( \boldsymbol{\approx 115 \, \text{s}} \) (rounded)
• Cage 2 time: \( \boldsymbol{\approx 117 \, \text{s}} \) (rounded)
• Cage 1 reaches the bottom first.

Problem 4:

Part 1: Acceleration

• Initial speed, \( u = 317 \, \text{km/hr} = 317 \times \frac{1000}{3600} \approx 88.06 \, \text{m/s} \)
• Final speed, \( v = 200 \, \text{km/hr} = 200 \times \frac{1000}{3600} \approx 55.56 \, \text{m/s} \)
• Time, \( t = 8.0 \, \text{s} \)

Using \( v = u + at \):
\[ a = \frac{v - u}{t} = \frac{55.56 - 88.06}{8.0} = \frac{-32.5}{8.0} \approx -4.06 \, \text{m/s}^2 \]

Part 2: Displacement
Using \( s = ut + \frac{1}{2}at^2 \):
\[ s = (88.06)(8.0) + 0.5(-4.06)(8.0)^2 \]
\[ s = 704.48 - 0.5(4.06)(64) \]
\[ s = 704.48 - 130.08 \]
\[ s \approx 574.4 \, \text{m} \]

Final Answer (Problem 4):

• Acceleration: \( \boldsymbol{\approx -4.06 \, \text{m/s}^2} \) (deceleration)
• Displacement: \( \boldsymbol{\approx 574 \, \text{m}} \)

Problem 5:

Phase 1: Accelerating to max speed

• Initial speed, \( u = 0 \, \text{m/s} \)
• Acceleration, \( a = 0.800 \, \text{m/s}^2 \)
• Final speed, \( v = 3.06 \, \text{m/s} \)

Time to accelerate (\( t_1 \)):
\[ v = u + at_1 \implies t_1 = \frac{v - u}{a} = \frac{3.06 - 0}{0.800} = 3.825 \, \text{s} \]

Distance during acceleration (\( s_1 \)):
\[ s_1 = ut_1 + \frac{1}{2}at_1^2 = 0 + 0.5(0.800)(3.825)^2 \approx 5.86 \, \text{m} \]

Phase 2: Constant speed

• Time, \( t_2 = 5.00 \, \text{s} \)
• Speed, \( v = 3.06 \, \text{m/s} \)

Distance (\( s_2 \)):
\[ s_2 = v \times t_2 = 3.06 \times 5.00 = 15.3 \, \text{m} \]

Total distance:
\[ s_{\text{total}} = s_1 + s_2 = 5.86 + 15.3 = 21.16 \, \text{m} \]

Final Answer (Problem 5):

The total distance is \( \boldsymbol{\approx 21.2 \, \text{m}} \) (rounded).

Problem 6:

Part 1: Time to reach max speed

• Initial speed, \( u = 0 \, \text{m/s} \)
• Acceleration, \( a = 4.00 \, \text{m/s}^2 \)
• Final speed, \( v = 350 \, \text{km/hr} = 350 \times \frac{1000}{3600} \approx 97.22 \, \text{m/s} \)

Using \( v = u + at \):
\[ t = \frac{v - u}{a} = \frac{97.22 - 0}{4.00} \approx 24.3 \, \text{s} \]

Part 2: Distance traveled
Using \( s = ut + \frac{1}{2}at^2 \):
\[ s = 0 + 0.5(4.00)(24.3)^2 \]
\[ s = 2.00 \times 590.49 \]
\[ s \approx 1181 \, \text{m} \]

Final Answer (Problem 6):

• Time: \( \boldsymbol{\approx 24.3 \, \text{s}} \)
• Distance: \( \boldsymbol{\approx 1180 \, \text{m}} \) (rounded)

Problem 7:

Given:

• Acceleration, \( a = 2.67 \, \text{m/s}^2 \)
• Time, \( t = 15.0 \, \text{s} \)
• Displacement, \( s = 600 \, \text{m} \)
• Initial speed, \( u = ? \)

Using \( s = ut + \frac{1}{2}at^2 \):

Step 1: Rearrange for \( u \)

\[ ut = s - \frac{1}{2}at^2 \]
\[ u = \frac{s - 0.5at^2}{t} \]

Step 2: Substitute values

\[ u = \frac{600 - 0.5(2.67)(15.0)^2}{15.0} \]
\[ u = \frac{600 - 0.5(2.67)(225)}{15.0} \]
\[ u = \frac{600 - 300.375}{15.0} \]
\[ u = \frac{299.625}{15.0} \approx 19.97 \, \text{m/s} \]
(Convert to \( \text{km/hr} \): \( 19.97 \times \frac{3600}{1000} \approx 71.9 \, \text{km/hr} \))

Final Answer (Problem 7):

The initial speed is approximately \( \boldsymbol{20.0 \, \text{m/s}} \) (or \( \boldsymbol{72.0 \, \text{km/hr}} \)).

Problem 8:

Given:

• Final speed, \( v = 300 \, \text{m/s} \)
• Acceleration, \( a = 7.20 \, \text{m/s}^2 \)
• Time, \( t = 25.0 \, \text{s} \)
• Initial speed, \( u = ? \)

Using \( v = u + at \):

Step 1: Rearrange for \( u \)

\[ u = v - at \]

Step 2: Substitute values

\[ u = 300 - (7.20)(25.0) \]
\[ u = 300 - 180 \]
\[ u = 120 \, \text{m/s} \]

Final Answer (Problem 8):

The initial speed is \( \boldsymbol{120 \, \text{m/s}} \).

Problem 9:

Given:

• Initial speed, \( u = 0 \, \text{m/s} \)
• Acceleration, \( a = 4.0 \, \text{m/s}^2 \)
• Final speed, \( v = 965 \, \text{km/hr} = 965 \times \frac{1000}{3600} \approx 268.06 \, \text{m/s} \)

Using \( v^2 = u^2 + 2as \):

Step 1: Rearrange for \( s \)

\[ s = \frac{v^2 - u^2}{2a} \]

Step 2: Substitute values

\[ s = \frac{(268.06)^2 - 0}{2(4.0)} \]
\[ s = \frac{71856.16}{8.0} \approx 8982 \, \text{m} \]
(Convert to \( \text{km} \): \( 8982 \, \text{m} = 8.98 \, \text{km} \))

Final Answer (Problem 9):

The distance is approximately \( \boldsymbol{9.0 \, \text{km}} \) (rounded).

Problem 10:

Given:

• Cruising speed, \( v_{\text{cruise}} = 2.30 \times 10^3 \, \text{km/hr} \)
• Landing speed, \( v = 20.0\% \) of cruising speed:

\( v = 0.20 \times 2.30 \times 10^3 = 460 \, \text{km/hr} = 460 \times \frac{1000}{3600} \approx 127.78 \, \text{m/s} \)

• Acceleration, \( a = -5.80 \, \text{m/s}^2 \) (deceleration)
• Final speed, \( v_f = 0 \, \text{m/s} \)

Using \( v_f^2 = v^2 + 2as \):

Step 1: Rearrange for \( s \)

\[ s = \frac{v_f^2 - v^2}{2a} \]

Step 2: Substitute values

\[ s = \frac{0 - (127.78)^2}{2(-5.80)} \]
\[ s = \frac{-16328.73}{-11.6} \approx 1407.6 \, \text{m} \]

Final Answer (Problem 10):

The distance traveled is approximately \( \boldsymbol{1410 \, \text{m}} \) (rounded).