Question

In: Physics

*Relativity and time - 1 The distance between New York and Los Angeles is about 5,000...

*Relativity and time - 1

The distance between New York and Los Angeles is about 5,000 km. A jet, traveling at a speed of 769 km/hour, leaves New York. Calculate the time elapse for the plane to reach the Los Angeles airport. Now calculate the difference in the time that has elapsed for an observor on earth and a passenger on the plane? Give your answers in nanoseconds (ns)

Solutions

Expert Solution

Given, d = 5000 km = 5 x 10⁶ m;

Speed of jet = v = 769 km/hr

For the passenger, Distance will contract to d^', where,

As v is very small as compared to c, we can approximate to using binomial theorem.

$\lambda = \left ( 1+ 2.535\times10^{-13})$

Again approximating,

For the passenger let time taken be t_p;

secs

For the observer let time taken be to;

secs

Difference in time elapsed =

= = secs

= 5.934 ns.

genius_generous answered 1 year ago

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