Question

An ant crawled from a hole to the food. For the first 6 minutes,it travelled at an average speed of 15m/min. For The next 9 minutes it travelles at an average speed of 10m/min what was the average speed of the whole journey?

Answer 1

v = 12 m/min

Step-by-step explanation:

• By definition, the average velocity is the rate of change of the position with respect to time, as follows:

`v_(avg) =(x_(f) -x_(o) )/(t_(f)-t_(o) ) (1)`

• Choosing x₀ = 0 and t₀ =0, (1) reduces to :

`v_(avg) = (x_(f) )/(t_(f) ) (2)`

• From the givens, we have:

tif = 6 min + 9 min = 15 min

• In order to get xf, we know that during the first part, vavg = 15 m/min, so solving for xf:

`x_(f1) = v_(avg1)* t_(1) = 15 m/min * 6 min = 90 m (3)`

• For the following 9 min, we know that the average speed was 10m/min, so the distance traveled during the second part of the trip was simply:

`x_(f2) = v_(avg2) *t_(2) = 10m/min * 9 min = 90 m (4)`

• Adding (3) and (4):

`x_(f) = 90 m + 90 m = 180 m (5)`

• Replacing xf and tif in (2), we finally get:

`v_(avg) =(x_(f) )/(t_(f)) =(180m)/(15 min) = 12 m/min (6)`