Velocity, time, and displacement

[sdoi]

Homework Statement

Using graphical methods as much as possible, answer the following questions:
(a) What is the displacement of the object at t=6.0s?
(b) What is the average velocity of the object between t=0.00s and t=6.0s?
(c) What is the average speed of the object between t=0.0s and t=6.0s?

Homework Equations

1/2(rise/run)
Vav= (delta)d/(delta)t

The Attempt at a Solution

(a)
=1/2(10/1)
(delta)d=5m

(b)
Vav= (delta)d/(delta)t
= 5m/6.0s
= 0.83m/sDoes this look right?

 

[NascentOxygen]

You should show your working in a way that allows us to see how you figured it out.
Have you added up a lot of areas to find displacement? If not, then (a) is wrong.

 

[grzz]

re (a)
The direction of the displacement is important. Work out distance in one direction and distance in the opposite direction and find resultant displacement.

 

[sdoi]

For (a):
Displacement= area
1/2 (1.5s)(15m [south])
=11.5m[south]-5m[south]
=6.25m[south]

For (b):
Vav= (delta)d/t
=6.25m [south] / 6.0s
Vav=1.04m/s [south]

 

[NascentOxygen]

(a) Displacement = the area under the velocity vs. time graph (see attachment)

area above the v=0 line is positive, area below the v=0 line is negative.

Displacement at t=6 is determined as the area shown shaded. You can calculate it by counting squares and fractions of squares, or by computing areas of triangles you can see. (Preferably, use both methods, and this way you can check that your calculations agree.) If displacement were to equal zero, it means you are back to where you started from (at t=0); but it doesn't mean you've been nowhere.

In case there is any ambiguity, I have emphasised the x-axis (line v=0) in red.

(b) Suppose you jump in your car and drive north for 10 miles at 200 mph, then spin around and return to your starting point at again 200 mph. Your displacement is 0. Your average velocity is 0. But your average speed is 200 mph and you earn a speeding ticket.

So take careful note whether you are asked for average velocity or average speed.