Question

A car traveling at 35.2 m/s takes 28.0 minutes to travel a certain distance according to the driver's clock in the car. How long does the trip take according to an observer at rest on Earth? Hint: The following approximation is helpful:

[1 − x]^−1/2 ≈ 1 + 1/2x for x<<1.

28.0 min + ________s

Answer 1

According to special theory of relativity, the expression for time dilation is

t= t0/(1 -(v/c)2 )1/2 where, v-speed of the object, c-speed of light

given: v = 35.2 m/s

c = 3 x 108 m/s

t0 = 28.0 minute

since, v is very small compare to c,

1/(1 - (v/c)2 )1/2 = ( 1 -(-1/2)(v/c)2 )

=( 1 +(1/2)(v/c)2 )

t= t0/(1 - (v/c)2 )1/2

= t0 ( 1 +(1/2)(v/c)2 )

=t0 + t0(1/2)(v/c)2

= 28.0 + 28.0 x ½ x (35.2/ 3 x 108 )2

=28.0 + (14.0 x (11.73 x 10-8 )2 )

=28.0 minute + 19.25 x 10-14 minute

Answer: time taken for the trip according to an observer at rest on the earth,

t = 28.0 minute + 19.25 x 10-14 minute