Motion help

[Inside Out]

Please help .      1.1 9)c. Fiona rides her skateboard up a ramp. She begins with a speed of 8.0ms^-1 but slows down with a constant deceleration of 2.0m/s/s. She travels some distance up the ramp before coming to rest, then rolls down again. Ignoring air resistence and friction, calculate:  fionas velocity after 5.0seconds have elapsed.      Answer: 2.0m/s down


edit rohitpi: changed topic name

[Aurelian]

Please help .      1.1 9)c. Fiona rides her skateboard up a ramp. She begins with a speed of 8.0ms^-1 but slows down with a constant deceleration of 2.0m/s/s. She travels some distance up the ramp before coming to rest, then rolls down again. Ignoring air resistence and friction, calculate:  fionas velocity after 5.0seconds have elapsed.      Answer: 2.0m/s down

u = 8.0ms-1
v = ?
a = -2.0ms-2
t = 5.0s

v = u + at = 8.0ms-1 + -2.0ms-2*5.0s = 8.0ms-1 - 10ms-1 = -2.0ms-1 (or, 2.0ms-1 down).

The important thing is to just keep track of your directions and be consistent as to which sign (negative or positive) you ascribe to which direction (up or down). In the above, I said u = 8.0ms-1, and since the initial velocity is known to be "up", and it's decelerating, then the deceleration magnitude must be given as down.

[Lasercookie]

Well you know that this question will require one of the equations of constant acceleration.

These questions are probably best approached by listing what you know, and then find the appropriate equation
(in this case, I'm going to say that +ve indicates up, -ve indicates down)
I find drawing a picture of the situation also helps.

We're trying to find v (final velocity), so in this case we can use:

Final velocity is 2 m/s down.

[Inside Out]

I dont get why a=-2. Isnt the direction still up. What exactly does decelleration mean i though it just meant slowong down so the sign doesnt change in front of the acceleration

[Aurelian]

I dont get why a=-2. Isnt the direction still up. What exactly does decelleration mean i though it just meant slowong down so the sign doesnt change in front of the acceleration

Think of it in terms of forces. If something is going up, the only way to slow it down is to exert a force going down - that is, a 'negative' force, the observed result of which is a deceleration. A deceleration is actually just an acceleration in the opposite direction. Does that help?

[Lasercookie]

I dont get why a=-2. Isnt the direction still up. What exactly does decelleration mean i though it just meant slowong down so the sign doesnt change in front of the acceleration

Yep, deceleration is slowing down. I chose to set it to negative (which relates to what Aurelian was saying about directions and staying consistent) to represent that it's slowing down by 2 m/s every second. If it was positive, it'd mean that it was speeding up by 2 m/s every second.

If you look at what's going on with the numbers:
(where u=8m/s and t=5s)
If 'a' was positive, the final velocity increase. Clearly that's not what we want, as you're meant to be slowing down - the final velocity should be lower than that initial velocity.

If 'a' is negative, the final velocity would decrease. Which is what we want.

You could choose to have the positive sign represent slowing down (say that +ve represents 'down' and -ve represents 'up'), but I feel that having it as negative is more intuitive (but that's just me).

[Inside Out]

So does that mean if youthrow a tennis ball in the air the accelearation is -9.8 since the velocity is decreasing to 0 and when te ball goes back down he acceleration is 9.8 cause it is accelerating and the velocity is increasing. But how can this be when i thought acceleration was always constant.

[Aurelian]

So does that mean if youthrow a tennis ball in the air the accelearation is -9.8 since the velocity is decreasing to 0 and when te ball goes back down he acceleration is 9.8 cause it is accelerating and the velocity is increasing. But how can this be when i thought acceleration was always constant.

No you're correct in your final statement; acceleration due to gravity IS always constant. This is a common "trick" question in exams. The 'acceleration' on the ball is -9.8ms-2 *at all times*. But just remember that the sign (negative/positive) is an arbitrary choice - again, what matters is only that you are consistent.

[Inside Out]

But how can it be constant if its deccelerating when it goes up and accelerating hen it goes down wouldnt the sign have to change regardless of whay durection is positive.

[Aurelian]

But how can it be constant if its deccelerating when it goes up and accelerating hen it goes down wouldnt the sign have to change regardless of whay durection is positive.

The motion may not be constant, but that doesn't mean the acceleration isn't constant.

As the ball goes up, it gets slower and slower until eventually it stops and then starts getting faster in the other direction. However, getting faster going in the other direction (that is, falling back down), is really just an extension of the original phase of "getting slower". Why is this the case? Because going down is esentially just going up... negatively. So what's really happened to the speed of the ball is that it's gone from positive, to zero, to negative - e.g., 3ms-1, 0ms-1, -3ms-1 - and in this respect it "always decreased" and furthermore always decreased at the same rate. Thus, the acceleration remains constant both in magnitude and in sign.

[Lasercookie]

There's probably better videos to demonstrate this, but this might be helpful in visualising the situation:
http://www.youtube.com/watch?v=JoWwZa-HoGA

It shows a ball being thrown up, and then shows how the displacement, velocity and acceleration varies.

[Inside Out]

Ahhhhhhhh! I get it. That youtube link really helped thanks everyone