please help physics unit 1&2

[amun]

 A question if u guys can help  please :physics unit 1 & 2


A a battery- operated car travels 6.0 meter north in 2.4 s and then 6.0 metre south in 1.8s

Its a distance time graph right ?


What is the displacement of the car?
What is the average velocity of the car?
what is the distance travelled by the car?
What is the average speed  of the car?


im confused because they didn't give me a graph or anything and how would i work out displacement and all

[HERculina]

A a battery- operated car travels 6.0 meter north in 2.4 s and then 6.0 metre south in 1.8s 

Its a distance time graph right ?
Dont think u need to draw a graph.
     
What is the displacement of the car? Displacement = 0 as car returns to starting pt
What is the average velocity of the car? v = x/t = 0/4.2 = 0
what is the distance travelled by the car? distance = 6 + 6 = 12m
What is the average speed  of the car? speed = distance/time = 12/4.2 = 2.86 m/s



Hope that helped

[Lasercookie]

travels 6 metres north in first 2.4 seconds, travels 6 metres south

1. Displacement is a vector, you need to take directions into account
North is +ve, South is -ve


2.

3. Distance is a scalar:

4. Speed is a scalar metres per second

edit: beaten

[amun]

okay thanks guys that helped alot .

[amun]

can i ask you guys one more question 

A racing car has an acceleration of 12.0 ms -2  along a straight northerly  section of track.  If the velocity of the car is  24.0 ms-1  as it enters the straight section ,what  will the velocity  of the car be  after it has travelled  for 2.0 second along the track?


is the answer 24 m s-1

[Gotam]

I think the answer would be 48 ms-1
 if his acceleration is 12 ms-2 and his starting velocity is 24 ms-1 then as he enters the track at 24 ms-1 and rides for 2 seconds his final velocity would be 48 ms-1

We can use the formula v=u+at
v=? (what we're trying to find)
u=24 ms-1 ( initial velocity as he enters the track)
t=2.0s
a=12 ms-2

v= 24 + 12(2)
v=48 ms-1           

Regarding your answer of 24 ms-1 if he is accelerating and begins at 24 ms-1 then his final velocity after 2 seconds cannot be the same as his initial velocity because he is accelerating

I hope this helped

[amun]

okay can i ask two more questions . Thanks btw


What is the value in ms-2 north of an acceleration of 250 cm s-2 . Just dont get it at all

 A space rocket  travelleing horizontally  at 24 m s-1 fires it second stage  for 3.0 s .At the end of this time the velocity is 180 m s-1 .


a.What acceleration does the second stage produce ?
I said  a= v-u/t    a=180-24-3.0 i got 172 m s-2

And i don't get the rest

b. What distance will the rocket travel while it is accelerating? 
c. what acceleration  would be needed to stop the rocket  in 12.0s
d. how far would the rocket travel in the  12.0 s while it was stopping ?



thanks i dont have answers because the teacher gave us a 5 worksheets :/

[Lasercookie]

I think you need to look at the six equations of constant acceleration again.
http://en.wikipedia.org/wiki/Equations_of_motion#Classic_version (note that they use 's', where I would normally use 'x')

To answer these questions first write down everything you know

E.g. for b you know
u=24m/s, t=3s, v=180ms, and then also a=172 m/s^2=(v-u)/t=(180-24)/3=156/3=52 (how'd you get 172?, you gave the same equation)

Now you can figure out which equation to apply.
It'll be the one where you can put in those values, but also has an 'x' in it.
In this case it might be or or

Try not to mindlessly plug and chug too much (though that's all VCE Physics is), keep some thought about what each value means and try to picture the situation. With that in mind, you should be able to answer most questions correctly.

About this:

What is the value in ms-2 north of an acceleration of 250 cm s-2 . Just dont get it at all

I also have no idea what this question is saying. Is it asking what's the magnitude of the north component of the vector? Was there any more information, i.e. does it give a direction for the a = 250cm/s^2?

[amun]

No their was not anymore information.


I have a question .

forces of 12.0 and.16.0 newton act at right angles on a mass of 4.0 kg. what is the accelration produced by the forces if the mass is placed.

a) on smooth surface
b) on a rough surface where there is a fri tional force of 2.0 opposing the motion?

it says use pythagoras but i dont get how u could use pythagoras to find the accelration

[amun]

And i also dong get how the acceleration between the rough and smooth surface except the fact that the rough surface has ra frictional force of 2.N

[b^3]

So use vector addition to find the resultant force. Acceleration and force are related F=ma, so the acceleration will be that resultant force divded by the mass, i.e. 4kg
That will work for the smooth surface.
For the rough surface, the friction force will oppose the motion, so it will be in the opposite direction so we take it away from the resultant force. From that you should be able to use the above methods to get acceleartion.

e.g. for the first one
(resultant force)²=12²+16²
resultant force=root(400)
=20N
use trig ratios to find the angle

[Lasercookie]

Forces are vectors (they have a magnitude AND a direction).
At right angles means at 90 degrees to each other.
Draw a horizontal line and say that is 12N
Draw a vertical line perpendicular to that line and say that is 16N (of course, you could switch the forces around, it doesn't matter).
The line connecting them is the resultant force. Since the resultant force is the hypotenuse, you can use pythagoras to find out the magnitude. You should be okay with it now?

Here's a diagram: