Jogger question

Question

A jogger starting their morning run accelerates from standstill to their steady jogging pace of 8 km h^-1. They ready a speed of 8 km h^-1, 5s after starting. How long does it take the jogger to reach the end of their 20m long driveway?

Answer 1

Solving any physics problem is a process. The first thing that needs to be done is to get the entire problem into the same units. In this problem, the speed is given in km/hr, time in seconds, and the distance in meters.

Step 1: Convert km/hr to m/s:

(8 km/hr)(1000 m/km)(1 hr/3600 s) = 2.22 m/s (or 2 m/s)

Step 2: determine the acceleration:

a = (v - v₀) / (t - t₀) = v / t = (2.2 m/s) / (5 s) = 0.44 m/s² (since v₀ = 0 and t₀ = 0) - NOTE, in this case, it isn't really necessary to put in the numbers

Step 3: determine how far the jogger travels while accelerating:

Using the general motion equation for distance: x = x₀ + v₀ t + (1/2) a t², with x₀ and v₀ both equal to zero, get: x = (1/2) a t² = (v t)/2 = [(2.2 m/s)(5 s)] / 2 = 5.5 m

Step 4: determine how much farther the jogger must go to reach the end of the driveway:

L - x = d, where L is the length of the driveway, x is the distance traveled while accelerating, and d is the distance traveled at constant speed. L - x = 20 m - 4.5 m = 14.5 m

Step 5: determine the time it takes to travel 14.5 m at 2.2 m/s:

using the general motion equation again, where now d_o = 0, a = 0, v₀ = 2.2 m/s, and d = 14.5 m, rearrange and solve for t:

t = d / v₀ = (14.5 m) / (2.2 m/s) =6.6 s (6.5 s - depending how many significant figures are carried)

Step 6: add the time accelerating, 5 s (given), to the time at constant speed, 6.6 s (calculated) 5 + 6.6 = 11.6 s (NOTE: depending where and when you round your answer may vary slightly)