- #1
RoryScanlan
- 11
- 0
Very simple question -- Average speed for a journey
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.
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
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 traveled 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 traveling away from the starting position we could say her velocity is positive and would be∶ Distance/Time=+Velocity and, If Tyler was traveling 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? I am unsure on how to explain velocity.
Homework Statement
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.
Homework 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
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 traveled 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 traveling away from the starting position we could say her velocity is positive and would be∶ Distance/Time=+Velocity and, If Tyler was traveling 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? I am unsure on how to explain velocity.