91影视

Write the expression for the time dilation.

$\mathit{纬}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}$ 鈥︹�� (1)

Here, v is the speed of a twin, and c is the speed of light.

(a)

Substitute $v=\frac{4}{3}c$in equation (1).

$\begin{array}{c}纬=\frac{1}{\sqrt{1鈭�\frac{{\left(\frac{4}{5}c\right)}^2}{c^2}}}\\ 纬=\frac{1}{\sqrt{1鈭�\frac{16}{25}}}\\ 纬=\sqrt{\frac{25}{9}}\\ 纬=\frac{5}{3}\end{array}$

As the time duration of one twin between${21}^{\text{st}}$ birthday and ${39}^{\text{th}}$birthday is 18 years, the time elapsed with respect to her twin brother will be,

$\begin{array}{c}T=纬\left(18\text{鈥墆别补谤蝉}\right)\\ T=\left(\frac{5}{3}\right)脳\left(18\text{鈥墆别补谤蝉}\right)\\ T=30\text{鈥墆别补谤蝉}\end{array}$

Hence, the age of a twin brother will be,

$\begin{array}{c}\text{Ageofabrother}=\left(21\text{鈥墆别补谤蝉}\right)+\left(30\text{鈥墆别补谤蝉}\right)\\ \text{Ageofabrother}=51\text{鈥墆别补谤蝉}\end{array}$

Therefore, the age of a twin brother is $51\text{鈥墆别补谤蝉}$.

(b)

Write the equation to calculate the distance of star X.

$d=vt$ 鈥︹�� (2)

Here, t is the time taken to reach star X, which is given as:

$t=\frac{T}{2}$

Substitute$T=30\text{鈥墆别补谤蝉}$in the above expression.

$\begin{array}{c}t=\frac{30\text{鈥墆别补谤蝉}}{2}\\ t=15\text{鈥墆别补谤蝉}\end{array}$

Substitute$v=\frac{4}{5}c$ and$t=15\text{鈥墆别补谤蝉}$ in equation (2).

$\begin{array}{c}d=\left(\frac{4}{5}c\right)脳\left(15\text{鈥墆别补谤蝉}\right)\\ d=\left(4c\right)脳\left(3\text{鈥墆别补谤蝉}\right)\\ d=12\text{鈥塴颈驳丑迟测别补谤蝉}\end{array}$

Therefore, the distance of star X is $12\text{鈥塴颈驳丑迟测别补谤蝉}$.

(c)

In frame S, write the x and t coordinates.

$\begin{array}{c}x=12\text{鈥塴颈驳丑迟测别补谤蝉}=12c\text{鈥墆别补谤蝉}\\ t=15\text{鈥墆别补谤蝉}\end{array}$

Hence, the coordinates will be,

$(x,t)=(12c,15)\text{鈥墆别补谤蝉}$

Therefore, the coordinates $(x,t)$of the jump is $(12c,15)\text{鈥墆别补谤蝉}$.

(d)

The twin got on at the origin in $\stackrel{炉}{S}$, and at the time of walking along the road on $\stackrel{炉}{S}$, she鈥檚 still at the origin, so, the $\stackrel{炉}{x}$coordinate will be,

$\stackrel{炉}{x}=0$

Write the equation to calculate the$\stackrel{炉}{t}$coordinate.

$\stackrel{炉}{t}=\frac{t}{纬}$

Substitute $t=15\text{鈥墆别补谤蝉}$and $纬=\frac{5}{3}$in the above equation.

$\begin{array}{c}\stackrel{炉}{t}=\frac{15}{\left(\frac{5}{3}\right)}\\ \stackrel{炉}{t}=15脳\frac{3}{5}\\ \stackrel{炉}{t}=9\text{鈥墆别补谤蝉}\end{array}$

Hence, the coordinates will be,

$(\stackrel{炉}{x},\stackrel{炉}{t})=(0,9)\text{鈥墆别补谤蝉}$

Therefore, the coordinates$(\stackrel{炉}{x},\stackrel{炉}{t})$ of the jump in$\stackrel{炉}{S}$ is$(0,9)\text{鈥墆别补谤蝉}$ .

(e)

Write the equation for the $\tilde{x}$coordinate in the frame $\tilde{S}$.

$\tilde{x}=纬(x+vt)$

Substitute $12c=x$, $v=\frac{4}{5}c$, $纬=\frac{5}{3}$and $t=15$in the above equation.

$\begin{array}{c}\tilde{x}=\frac{5}{3}\left(12c+\left(\frac{4}{5}c\right)\left(15\right)\right)\\ \tilde{x}=\frac{5}{3}(12c+12c)\\ \tilde{x}=\frac{5}{3}\left(24c\right)\\ \tilde{x}=40\text{鈥塴颈驳丑迟测别补谤蝉}\end{array}$

Write the equation for the$\tilde{t}$coordinate in frame $\tilde{S}$.

$\tilde{t}=纬\left(t+\left(\frac{v}{c^2}\right)x\right)$

Substitute $12c=x$, $v=\frac{4}{5}c$, $纬=\frac{5}{3}$and $t=15$in the above equation.

$\begin{array}{c}\tilde{t}=\frac{5}{3}\left(15+\left(\frac{\left(\frac{4}{5}c\right)}{c^2}\right)\left(12c\right)\right)\\ \tilde{t}=\frac{5}{3}\left(15+\left(\frac{4}{5}c脳\frac{1}{c^2}\right)\left(12c\right)\right)\\ \tilde{t}=\frac{5}{3}\left(15+\frac{48}{5}\right)\end{array}$

On further solving,

$\begin{array}{c}\tilde{t}=\frac{5}{3}\left(\frac{75+48}{5}\right)\\ \tilde{t}=\frac{5}{3}\left(\frac{123}{5}\right)\\ \tilde{t}=41\text{鈥墆别补谤蝉}\end{array}$

Hence, the coordinates will be,

$(\tilde{x},\tilde{t})=(40c,41)\text{鈥墆别补谤蝉}$

Therefore, the coordinates $(\tilde{x},\tilde{t})$of the jump in$\tilde{S}$ is$(40c,41)\text{鈥墆别补谤蝉}$ .

(f)

Using the coordinate time from$9\text{鈥墆别补谤蝉}$ to $41\text{鈥墆别补谤蝉}$, the return trip takes$32\text{鈥墆别补谤蝉}$ so, the timing in her watch when she gets home will be,

$\begin{array}{c}\text{Time}\left(\text{whenshegetshome}\right)=41\text{鈥墆别补谤蝉}+9\text{鈥墆别补谤蝉}\\ \text{Time}\left(\text{whenshegetshome}\right)=50\text{鈥墆别补谤蝉}\end{array}$

Therefore, the reading of a watch when she gets home is $50\text{鈥墆别补谤蝉}$.

(g)

(i)Just before she makes the jump:

Calculate the age of a brother just before she makes the jump.

$\begin{array}{c}t=\frac{\stackrel{炉}{t}}{纬}\\ t=\frac{9\text{鈥墆别补谤蝉}}{\left(\frac{5}{3}\right)}\\ t=\left(9\text{鈥墆别补谤蝉}\right)脳\left(\frac{3}{5}\right)\\ t=5.4\text{鈥墆别补谤蝉}\end{array}$

As he started walking on their$21{\text{鈥�}}^{\text{st}}$birthday, i.e., at the age of $21\text{鈥墆别补谤蝉}$, the age of a brother will be,

$\begin{array}{c}\text{Ageofabrother}=\left(21\text{鈥墆别补谤蝉}\right)+\left(5.4\text{鈥墆别补谤蝉}\right)鈯�\\ \text{Ageofabrother}=26.4\text{鈥墆别补谤蝉}\end{array}$

(ii)Just after she makes the jump:

Calculate the age of a brother just after she makes the jump.

$\begin{array}{c}t=\frac{\stackrel{炉}{t}}{纬}\\ t=\frac{41\text{鈥墆别补谤蝉}}{\left(\frac{5}{3}\right)}\\ t=\left(41\text{鈥墆别补谤蝉}\right)脳\left(\frac{3}{5}\right)\\ t=24.6\text{鈥墆别补谤蝉}\end{array}$

As he started walking on their$21{\text{鈥�}}^{\text{st}}$ birthday, i.e., at the age of $21\text{鈥墆别补谤蝉}$, the age of a brother will be,

$\begin{array}{c}\text{Ageofabrother}=\left(21\text{鈥墆别补谤蝉}\right)+\left(24.6\text{鈥墆别补谤蝉}\right)\\ \text{Ageofabrother}=45.6\text{鈥墆别补谤蝉}\end{array}$

Therefore, (i) the age of a brother just before she makes the jump is$\text{26.4鈥墆别补谤蝉}$ and (ii) the age of a brother just after she makes the jump is $45.6\text{鈥墆别补谤蝉}$.

(h)

It will take another $5.4\text{鈥墆别补谤蝉}$of earth time for the return. Hence, the brother鈥檚 age will be,

$\begin{array}{c}\text{Ageofabrother}=\left(45.6\text{鈥墆别补谤蝉}\right)+\left(5.4\text{鈥墆别补谤蝉}\right)\\ \text{Ageofabrother}=51\text{鈥墆别补谤蝉}\end{array}$

On comparing the above result to part (a), the age of a brother is exactly the same.

Therefore, the time required to return back is$5.4\text{鈥墆别补谤蝉}$ and the expected age of a brother at their reunion is $51\text{鈥墆别补谤蝉}$.