Displacement Calculation for a Moving Cyclist

[FlipMC]

Homework Statement

A cyclist is traveling with an average velocity of 5.9 m/s[W]. What will be his displacement after 1.2 h?

Homework Equations

d = v1▲t+½a▲t^2 ?

The Attempt at a Solution

Vavg = 5.9 m/s
▲t = 1.2 x 60 = 72 m/s

What I did was use d = v x t and I got 424m/s, which is 1, 526.4 km/h. But that's way off from my the textbook's answer which is, 26km[W]..

It sounds so easy but honestly, I'm going crazy trying to figure it out LOL .

 

[rl.bhat]

Cyclist is traveling with uniform velocity.
So the displacement = velocity*time.
Convert 1.2 hours to seconds

 

[Kalvarin]

You got to get everything in the same units. We have velocity in m/s so we want time in seconds.

1.2 hours = 72 minutes = 4320 seconds

Then just use:

Distance = Velocity*Time

 

[FlipMC]

Thanks guys.
So I did D = 5.9m/s x 4320 which equaled to 25, 488m/s?
Then, to change it back to km/h, times 3.6 then it equaled 91, 756.8?!
What am I doing wrong? :(

 

[rl.bhat]

In the problem displacement is asked, not the velocity.

 

[FlipMC]

Hm, right.. AMG, I still don't get this :'(