how do i find average velocity?
total distance divided by total time :)
\[v _{ave}= \Delta displacement/\Delta time\]
average velocity is variation of speed divided by variation of time
Use displacement, not distance
yep
So for example, if you went 10m East in 5 seconds, your average velocity would be 10m/5s = 2m/s
What is the average velocity of the car between t = 0 and t = 50 s?
how much that car has moved in meters?
We need to know the change in position
thats all i got so far
its not 50/0 lol
??
stuck?
wtf! I lost my answers I gave you..weird!
try this again
ok
ok, the car is moving on a line. Remember the y-axis is the cars velocity NOT it's position. Since the velocity is always positive, the car is always moving in the same direction and never turns around and goes back (if it did, we would have negative values for the velocity). The change in position then is just the total distance travelled which is 2km=2000m according to your screenshot. The average velocity is 2000m/50s=40m/s
says 40 is wrong
it says this Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully
???? does it want the answer in m/s?
Are we talking average velocity here or average acceleration?
Determine the average acceleration of the object in the following time interval t = 7.50 s to t = 22.5 s.
m/s^2
Oh! That's a totally different question. I solved the average velocity for the entire trip.
lol oh lord
Ok, we need to divide the change in velocity by the change in time. Time is easy 22.5-7.5=15.0s. It looks to me like the velocity at 7.5s is 20m/s (these can be difficult to estimate from a graph on a computer screen though). The final velocity is 50m/s. The change is therefore 30m/s. Average acceleration is (30m/s)(15s)=2m/s^2
sorry I didn't put in the division sign above hehe
Hope this is correct, I got to go! If it is wrong my estimate of the velocity at 7.5s could be off!
so u think its 50?
im sorry...30
its not 50 or 30 lol
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