A marathon runner completes a 42.188 km course in 2 h, 27 min, and 11 s. There is an uncertainty of 27 m in the distance traveled and an uncertainty of 1 s in the elapsed time.

Question

A marathon runner completes a 42.188 km course in 2 h, 27 min, and 11 s. There is an uncertainty of 27 m in the distance traveled and an uncertainty of 1 s in the elapsed time. (Give all answers to the appropriate number of significant figures. Note that uncertainties should be reported to one significant figure and percent uncertainties should be reported to two significant figures.)

a) Calculate the percent uncertainty in the distance.

= ?%

b) Calculate the percent uncertainty in the elapsed time. 

= ?%

c) What is the average speed in meters per second? 

= ?m/s

d) What is the uncertainty in the average speed? 

= ?m/s

Answer 1

(a). 42.188 km = 42188m, so % uncertainty in distance = (27 / 42188) x 100

(b) 2 hr 27 mins 11 s = 2 x 3600 + 27 x 60 + 11 s = 8831s. So (1/8831) x 100

(c) Mean speed in m/s is 42188 / 8831 = 4.7777 m/s

(d) Max and minimum distances are 42188 + 27, 42188 - 27

Max and minimum times are 8831+1, 8831-1 s

So max average speed = (42188 + 27) / (8831-1) = 4.7809 m/s

Min average speed = (42188 - 27) / (8831 + 1) = 4.7737

So range of average speeds is [ 4.7737, 4.7809 ] so uncertainty = 4.7809 - 4.7737 = 0.0072 m/s.