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

Answer:

v = 12 m/min

Explanation:

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

        $v_{avg} =\frac{x_{f} -x_{o} }{t_{f}-t_{o} } (1)$

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

        $v_{avg} = \frac{x_{f} }{t_{f} } (2)$

• 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:

       $v_{avg} =\frac{x_{f} }{t_{f}} =\frac{180m}{15 min} = 12 m/min (6)$