# Very simple question - Average speed for a journey

1. Nov 22, 2013

### RoryScanlan

Very simple question -- Average speed for a journey

1. The problem statement, all variables and given/known data
Tyler travels a distance (in a straight line) of 18km in 30minutes.

iii)Tyler Returns home following the same route in the same amount of time. Write down her average speed and velocity for this journey, explaining why they are not the same.

2. Relevant equations

The previous question i answered:

Tyler Travels : 18km in 30 minutes
18km=18 ×1000=18000m
30 minutes =30 ×60=1800seconds
Speed= 18000/1800=10ms^(-1)
3600 seconds = 1 hour ∴10 ×3600=36000meters per hour
36000 ÷1000=36km/h
Tyler’s average speed = 10ms-1 or 36km/h

3. The attempt at a solution

If Tyler returns home using the same route and in the same amount of time. Tyler’s average speed would be 10ms-1 or 36km/h. We know this because her route is identical and she first travelled in a straight line, therefore her return journey must be a straight line and the distance on the return journey must be 18000m and she returned in the same amount of time, i.e. 30 minutes or 1800 seconds. Therefore her speed equation would be identical, i.e.
Speed=18000/1800=10ms^(-1) or 36km/h

Tyler’s velocity however would differ on the return journey; Velocity is a vector it takes speed into account but also applies a direction. Therefore if Tyler was travelling away from the starting position we could say her velocity is positive and would be∶ Distance/Time=+Velocity and, If Tyler was travelling toward the starting point her velocity would be Distance/Time= -Velocity . We now assign it a negative value to show that the movement is in an opposite direction. Therefore on a return journey her velocity we could say is -10ms-1 or -36km/h.

Is this an acceptable answer? Im unsure on how to explain velocity.

2. Nov 22, 2013

### Staff: Mentor

When the problem statement said "average speed and velocity", it really meant to say "average speed and average velocity" or "average- speed and velocity". So, what was her average velocity?

3. Nov 22, 2013

### RoryScanlan

I think its talking about the return journey so if i think about a displacement over time graph it would positive 18000 down to 0 on the return journey for displacement therefore giving displacement of -18000, and total time of 1800 seconds. Therefore average velocity for the return journey is -18000/1800= -10m/s?

Last edited: Nov 22, 2013
4. Nov 22, 2013

### Staff: Mentor

If she returns to the same spot that she started from, the displacement vector from her starting point to her end point is zero. So, what is her average velocity (vector)?

5. Nov 22, 2013

### HallsofIvy

You don't really need to look at "18 km" or "1800 m". She started from a given point then, through whatever distance she traveled or what ever turns she may have taken, she returned to the same point. Her vectorial displacement is the 0 vector.

6. Nov 22, 2013

### RoryScanlan

Ok thanks, i was unsure if the question was asking about the whole journey or just the return trip. Having 0 displacement from the origin makes sense tho, ie displacement/time or 0/1800 = 0ms
Thanks guys :).

Last edited: Nov 22, 2013