Question

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

(a) How fast must the vehicle move (in units of c) so that its length measured by an earthbound observer is 49 m?

(b) Refer to part (a). If the vehicle carries passengers a distance of 6,000 km, by how much will the travel time measured by a clock on the vehicle differ from the time measured by an earth-based clock? In other words, what is Δt vehicle − Δt earth ?

Answer 1

a)

`5.97* 10^(7)` ms⁻¹

b)

`0.0021` sec

Step-by-step 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 ( (v)/(c) \right )^(2)}`

`49 = 50 \sqrt{1 - \left ( (v)/(3* 10^(8)) \right )^(2)}`

`0.98 = \sqrt{1 - \left ( (v)/(3* 10^(8)) \right )^(2)}`

`0.9604 = 1 - \left ( (v)/(3* 10^(8)) \right )^(2)`

`0.396 = \left ( (v)/(3* 10^(8)) \right )^(2)`

`v = 5.97* 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 = (d)/(v)`

`t = (6* 10^(6))/(5.97* 10^(7))`

`t = 0.1006` sec

Time measured by a clock on the earth is given as

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

`0.1006 = t'\sqrt{1 - \left ( (5.97* 10^(7))/(3* 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