Question

An astronaut exercising on a treadmill maintains a pulse rate of 150 per minute. If he exercises for 1.00 h as measured by a clock on his spaceship, with a stride length of 1.00 m/s, while the ship travels with a speed of 0.900c relative to a ground station, what are (a) the pulse rate and (b) the distance walked as measured by someone at the ground station?

Short answer

(a) The pulse rate at ground station is $65.4\raisebox{1ex}{$pulse$}\!\left/ \!\raisebox{-1ex}{$min$}\right.$.

(b) The distance measured at ground station is$1569.2m$ .

Step-by-step solution

 Step 1: Identification of given data

The pulse rate of astronaut on treadmillis $r_o=150\raisebox{1ex}{$pulse$}\!\left/ \!\raisebox{-1ex}{$min$}\right.$

The duration for exercise for astronaut is $t=1h=3600s.$

The stride length is $u=1\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The speed of ship is $v=0.900c.$

The pulse rate is the frequency of heart rate in every minute. Each pulse is made from one compression and one expansion of heart.

 Step 2: Determination of pulse rate at ground station 

(a)

The pulse rate at ground station is given as

$r=r_o\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

Here, $c$ is the speed of light and its value is $3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Substitute all the values in the above equation.

$$\begin{gathered} r=\left(150\raisebox{1ex}{$pulse$}\!\left/ \!\raisebox{-1ex}{$min$}\right.\right)\sqrt{1-{\left(\frac{0.900c}{c}\right)}^2} \\ r=65.4\raisebox{1ex}{$pulse$}\!\left/ \!\raisebox{-1ex}{$min$}\right. \end{gathered}$$

Therefore, the pulse rate at ground station is $65.4\raisebox{1ex}{$pulse$}\!\left/ \!\raisebox{-1ex}{$min$}\right.$.

Determination distance measured at ground station

(b)

The distance measured at ground station is given as:

$d=u{t}_o\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

Substitute all the values in the above equation.

$d=\left(1\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)\left(3600s\right)\sqrt{1-{\left(\frac{0.900c}{c}\right)}^2}$

$d=1569.2m$

Therefore, the distance measured at ground station is$1569.2m.$ .