Question

A special vehicle of length 50 m is designed to take passengers at extremely high speeds between different places on earth.

Answer 1

Answer:

a)

$5.97\times 10^{7}$ ms⁻¹

b)

$0.0021$  sec

Explanation:

(a)

$L_{o}$ = Actual length of the vehicle = 50 m

$L$ = length measured by the earthbound observer = 49 m

$v$ = speed of the vehicle

$c$ = speed of light = 3 x 10⁸ ms⁻¹

Length measured by the earthbound observer is given as

$L = L_{o}\sqrt{1 - \left ( \frac{v}{c} \right )^{2}}$

$49 = 50 \sqrt{1 - \left ( \frac{v}{3\times 10^{8}} \right )^{2}}$

$0.98 = \sqrt{1 - \left ( \frac{v}{3\times 10^{8}} \right )^{2}}$

$0.9604 = 1 - \left ( \frac{v}{3\times 10^{8}} \right )^{2}$

$0.396 = \left ( \frac{v}{3\times 10^{8}} \right )^{2}$

$v = 5.97\times 10^{7}$ ms⁻¹

b)

$d$ = distance traveled = 6000 km = 6 x 10⁶ m

$t$ = time measured by a clock on the vehicle

$t'$ = time measured by a clock on the earth

Time measured by a clock on the vehicle  is given as

$t = \frac{d}{v}$

$t = \frac{6\times 10^{6}}{5.97\times 10^{7}}$

$t = 0.1006$ sec

Time measured by a clock on the earth is given as

$t = t'\sqrt{1 - \left ( \frac{v}{c} \right )^{2}}$

$0.1006 = t'\sqrt{1 - \left ( \frac{5.97\times 10^{7}}{3\times 10^{8}} \right )^{2}}$

$t' = 0.1027$ sec

$\Delta t$ = time difference

Time difference is given as

$\Delta t = t' - t$

$\Delta t = 0.1027 - 0.1006$

$\Delta t = 0.0021$

$\Delta t = 0.0021$  sec