Related questions with answers

Why is identifying the reference frame important in describing motion?

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Question

A spaceship at rest in a certain reference frame S is given a speed increment of 0.50c. Relative to its new rest frame, it is then given a further 0.50c increment. This process is continued until its speed with respect to its original frame S exceeds 0.999c. How many increments does this process require?

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When a particle is moving with speed $u^\prime$ in the positive $x^\prime$ direction in an inertial reference frame $S^\prime$ that itself is moving with speed $v$ parallel to the $x$ direction of a second inertial frame $S$ , the speed $u$ of the particle as measured in $S$ is:

u=\dfrac{u^\prime +v}{1+u^\prime v/c^2}

at each increment the speed of frame of the spaceship is $0.50c$ , relative to the new frame. The spaceship starts from rest, so at the first increment $u^\prime=0$ , and the speed of the frame is $v=0.50c$ , so:

u_1=\dfrac{0 +v}{1+0}=0.50c

in the second increment $u^\prime=u_1=0.50c$ and $v=0.50c$ , so:

u_2=\dfrac{0.50c +0.50c}{1+(0.50c)(0.50c)/c^2}=0.80c

in the third increment $u^\prime=u_2=0.80c$ and $v=0.50c$ , so:

u_3=\dfrac{0.80c +0.50c}{1+(0.80c)(0.50c)/c^2}=0.9286c

in the fourth increment $u^\prime=u_3=0.9286c$ and $v=0.50c$ , so:

u_4=\dfrac{0.9286c+0.50c}{1+(0.9286c)(0.50c)/c^2}=0.9756c

in the fifth increment $u^\prime=u_4=0.9756c$ and $v=0.50c$ , so:

u_5=\dfrac{0.9756c+0.50c}{1+(0.9756c)(0.50c)/c^2}=0.9918c

in the sixth increment $u^\prime=u_5=0.9918c$ and $v=0.50c$ , so:

u_6=\dfrac{0.9918c+0.50c}{1+(0.9918c)(0.50c)/c^2}=0.9973c

in the seventh increment $u^\prime=u_6=0.9973c$ and $v=0.50c$ , so:

u_7=\dfrac{0.9973c+0.50c}{1+(0.9973c)(0.50c)/c^2}=0.9990c

Thus, seven increments are needed.

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