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Gemiola [76]
3 years ago
5

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d

ecelerates at this same rate for the second half of the distance. If the stations are 3200m apart, find the time of travel (in seconds) between the two stations.
Physics
1 answer:
Minchanka [31]3 years ago
7 0

This problems a perfect application for this acceleration formula:

         Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:     1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half:    1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by  0.65 m/s²: 

                         T² = (1600 m) / (0.65 m/s²)

                         T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

                         T =  2 √(1600/0.65) = <em>99.2 seconds</em>  (rounded)


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The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  V_x'=0.8c\\V=0.6c\\m=4*10^5kg

  • Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

                  KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}

  • So, to V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
  • We have an expression from Lorents transformation for relativistic law of addition of velocities as,

                      V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V

  • Substituting values, we get,

          V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c

  • Thus, the KE will be,

              KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

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