- #1

- 96

- 0

******acceleration In One Dimension*****

Hi everyone, I need some major help right now, I dun understand some concepts, im a first time physics student. Can an object have a negative acceleration and be speeding up?

- Thread starter Faiza
- Start date

- #1

- 96

- 0

Hi everyone, I need some major help right now, I dun understand some concepts, im a first time physics student. Can an object have a negative acceleration and be speeding up?

- #2

- 7

- 1

Yeah, if it is going in the negative direction. E.g backwards

- #3

- 150

- 0

it is all based on your frame of reference. that means that if you call the direction that the object is accelerating in the negative direction, then it will have a negative acceleration and be increasing its speed.Faiza said:Hi everyone, I need some major help right now, I dun understand some concepts, im a first time physics student. Can an object have a negative acceleration and be speeding up?

frames of reference are important parts of setting up Physics problems. it is best to think about the problem before hand and set it up so that its final destination or the direction is is mostly moving in (the ground, point b, etc) is in a positive direction. it makes the problems much easier to do.

- #4

- 399

- 1

- #5

- 96

- 0

Acceleration means speeding up

I think lol im a first time physics student i dun know :(

- #6

- 7

- 1

- #7

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 132

(a pair I prefer to ac/dec).

- #8

- 1,010

- 1

negitive acceleration just means you are speeding up in the direction opposite to the one defined as positive. But you could I guess say that you were decelerating from -25 m/s^2 to -20 m/s^2

deceleration really can mean different things based upon how you use them. I generally think of it as something that means your speed is slowing down

aka closer to 0 m/s

- #9

- 96

- 0

straight down with a speed of 12.0 m/s. He throws another pebble straight

upward with the same speed so that it misses the edge of the bridge on the way

back down and falls into the river. For each stone find (a) the velocity as it

reaches the water and (b) the average velocity while it is in flight.

Note: Ignore the affects of air resistance.

(a) X= 20 m a=9.80m/s2 Vi=12.0 m/s Vf= ?????

Vf2 = Vi2 + 2a (ΔX)

Vf2 = (12.0 m/s)2 + 2(9.80 m/s2) (20 m)

Vf2 = (144 m2/s2) + (19.6 m/s2) (20 m)

Vf2 = (144 m2/s2) + (392 m2/s2)

Vf2 = (536 m2/s2)

Vf = 23.15167381 m/s

Vf = 23 m/s

Therefore the velocity of the stone as it reaches the water is approximately 23 m/s. Since the pebble was thrown straight upward with the same speed so that it misses the edge of the bridge on the way back down and falls into the river, I know that if the pebble is projected upwards in a vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity it has when it returns to the same height. That is, a pebble projected vertically with an upward velocity of +12 m/s will have a downward velocity of –12 m/s when it returns to that same height.

(b) *AVERAGE VELOCITY*= ΔX/Δt

STONE

Xf = 20 m Vxf = 23 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????

Vxf = Vxi + ax t Vavg= ΔX/Δt

23 m/s = 12.0 m/s + (9.80 m/s2) t Vavg= 20m/1.12244898 s

23 m/s – 12.o m/s = (9.80 m/s2) t Vavg= 17.81818181 m/s

11 m/s = (9.80 m/s2) t Vavg= 17.81 m/s

t = 11 m/s .

9.80 m/s2

t = 1.12244898 s

t = 1.12 s

(b) *AVERAGE VELOCITY*= ΔX/Δt

PEBBLE

Xf = 20 m Vxf = 12.0 m/s Vxi=12.0 m/s ax=9.80m/s2 Xi = 0 m t=????

Vxf = Vxi + ax t

12.0 m/s = 12.0 m/s + (9.80 m/s2) t

12.0 m/s – 12.0 m/s= (9.80 m/s2) t

0 m/s = (9.80 m/s2) t

0 m/s = t

9.80 m/s2

t = 0 m/s

CHECK MY ASNWER???

- #10

- 96

- 0

How can you tell if the object is speeding up (acceleration) or slowing down (deceleration)? Speeding up means that the magnitude (the value) of the velocity is increasing. For instance, an object with a velocity changing from +3 m/s to + 9 m/s is speeding up. Similarly, an object with a velocity changing from -3 m/s to -9 m/s is also speeding up. In each case, the magnitude of the velocity (the number itself, not the sign or direction) is increasing; the speed is getting larger.

Given this fact, an object is speeding up if the line on a velocity-time graph is changing from a location near the 0-velocity point to a location further away from the 0-velocity point. That is, if the line is moving away from the x-axis (the 0-velocity point), then the object is speeding up. Conversely, if the line is moving towards the x-axis, the object is slowing down.

- #11

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 9,970

- 132

[tex]\vec{v}\cdot\vec{a}\geq{0}[/tex]

then the object "accelerates", otherwise it "decelerates".

([tex]\vec{v}[/tex] velocity, [tex]\vec{a}[/tex] rate of change of velocity (the proper meaning of acceleration)

- #12

- 5,755

- 1,051

deceleration, "slows you down" , "reduces your speed"

symbolically (and without reference to coordinates): [tex]\vec a\cdot \vec v < 0[/tex].

Proof:

[tex]0>\vec a\cdot \vec v =\frac{d \vec v}{dt}\cdot \vec v=\frac{1}{2} \frac{d (\vec v \cdot \vec v)}{dt}=\frac{1}{2}\frac{d (v^2)}{dt}=

\frac{1}{2}\frac{d}{dt}((\text{speed})^2)[/tex]

So,

[tex]0>\frac{d}{dt}(\text{speed})[/tex]

- Replies
- 4

- Views
- 1K

- Replies
- 2

- Views
- 6K

- Replies
- 2

- Views
- 5K

- Last Post

- Replies
- 5

- Views
- 27K

- Replies
- 4

- Views
- 16K

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 2

- Views
- 7K

- Replies
- 4

- Views
- 3K

- Replies
- 3

- Views
- 7K

- Replies
- 3

- Views
- 2K