kinematics?

[allstar]

can someone please help with this question? im stuck on the subjected to a constant acceleration in the opposite direction, shouldn't it be starting from below the x axis at that point?

[Zealous]

can someone please help with this question? im stuck on the subjected to a constant acceleration in the opposite direction, shouldn't it be starting from below the x axis at that point?

The acceleration is constant in the opposite direction: this means that the acceleration will be a constant negative 5m/s² from t=10s. But, this does not mean the velocity will suddenly become negative.

The acceleration is how the velocity changes with time, if the acceleration is negative, then velocity is decreasing, so the velocity will be reduced from 20m/s to 0m/s in 4 seconds. So for your velocity-time graph, take the section after t=10 seconds and flip it above the time axis.

[allstar]

cause I know that when the velocity changes direction, is becomes on the other side of the x - axis but why doesn't this apply to acceleration, how would I show that the object accelerated in the opposite direction, doesn't the negative gradient only show that it is decelerating, not necessary in the opposite direction?

[keltingmeith]

I want you to run at a speed of 5m/s one way, and then in an instant (without stopping) suddenly be running 5 m/s in the opposite direction.

Hopefully you have realised that this isn't something one can do - and so we meet our first rule of thumb. Velocity-time graphs should always be continuous, it is impossible (AFAIK, and I doubt an alternative case would show up in high school mechanics) for something to suddenly change velocities. The same rule should apply for acceleration (note: if acceleration/velocity time curves aren't continuous, we can't find the resulting velocity/displacement (respectively). Any clues as to why?)

If the acceleration happens in the direction opposite to the current direction of motion, then the object will experience a net DECELERATION. Note, this object will not stop decelerating when velocity is 0, because it returns to its original POSITION. You'll need to use integration/area formulae to figure it at what time it actually stops.

[allstar]

but cant it decelerate while travelling in the original direction? how do we differentiate between the two scenarios?

[keltingmeith]

It can decelerate in the original direction.

What you really need to get your head around is the fact that these are vector quantities, not just numbers. You really need to think about directions when doing kinematics problems. However, you also need to realise that acceleration and velocity, while linked, don't have to be in the same direction. To figure our your scenarios, make judgements based on the situation at hand and what way each different vector is pointing. In specialist, we only consider two directions - positive and negative. Of course, a particle can travel in neither of these directions, at which point it has no acceleration or no velocity. The same ideas can be extended to displacement.

If your velocity is positive and your acceleration is positive, your velocity will stay positive and increase.
If your velocity is positive and your acceleration is negative, your velocity will decrease to zero and then continue into the negatives.
If your velocity is negative and your acceleration is positive, your velocity will increase to zero and then continue into the positives.
If your velocity is negative and your acceleration is negative, your velocity will stay negative and decrease.

Now, in this case let's consider how acceleration is changing.

For this case, acceleration is zero. Then, we start to decelerate. Since acceleration is zero, decelerating will cause the acceleration to take a negative direction.

(note: apologies for ramblings, but I am currently sick and under the influence of way too many drugs to think properly. By the sounds of what I'm reading, you may need to go through your definitions a bit more and make sure that you properly acquainted with the idea of what a vector quantity is and how to treat it if you're not using the vector format you're used to [ie. i-j-k form])

[allstar]

but it says subjected to a constant acceleration?

[keltingmeith]

Yes - it's subjected to a constant accelerate IN THE OPPOSITE DIRECTION.

I repeat, these are vector quantities. If you accelerate something in the direction opposite to its current motion, it will decelerate. Directions are so important in this unit, you cannot afford to not understand that a force can have drastically different effects depending on which direction it is applied.

[allstar]

so when it says accelerating in the opposite direction, I assume it is decelerating and that if is continues to decelerate shouldn't it come to a rest at some point?

whereas when you think of accelerating in the opposite direction alone, can't it accelerate forever just in the opposite direction?

[keltingmeith]

Deceleration is just a term to define acceleration in the negative direction. If an object comes to rest, that doesn't mean it has stopped decelerating - rather, it can continue to decelerate such that it starts moving backwards.

The only time you should assume something stops decelerating when it comes to rest is when the question says, "x starts to decelerate until it comes to rest".

The issue with thinking of acceleration in the opposite direction is that you might think that because it's accelerating in the opposite direction, it must also be moving in the opposite direction. That's not the case, and until the object comes to rest, it will still be moving forwards, even though it's accelerating backwards. Hence why it's better to think of it as decelerating until it comes to rest, and then think of it as accelerating backwards (or just think of the whole thing as decelerating).

[allstar]

ok I think im starting to get it now! thank you

[allstar]

The maximum rate at which a bus can accelerate or decelerate is 2 m/s2. It has a
maximum speed of 60 km/h. Find the shortest time the bus can take to travel between
two bus stops 1 km apart on a straight stretch of road.

please

[Phenomenol]

The maximum rate at which a bus can accelerate or decelerate is 2 m/s2. It has a
maximum speed of 60 km/h. Find the shortest time the bus can take to travel between
two bus stops 1 km apart on a straight stretch of road.

please

First, 60 km/h = 60 * 5/18 m/s = 50/3 m/s (unit conversion)
Also, 1 km = 1000 m

We need to assume the bus starts from rest at stop A and and comes to a halt at stop B. So the bus's journey goes like:
- Accelerate to 60 km/h from A (as quick as possible)
- Maintain 60 km/h for a certain distance and time (for as long as possible)
- Decelerate from 60 km/h to rest and arrive at B (as quick as possible)

So, the problem is split as follows:
1. Find the minimum time taken to accelerate from rest to 60 km/h. Also find the distance covered.
2. Find the minimum time taken to decelerate from 60km/h to rest. (note: numerically the same, signs are flipped and cancel) Again find the distance covered.
3. Find the total distance covered during acceleration and deceleration. Subtract this from the known total distance (1km) to find the distance covered during which 60km/h is maintained
4. Find the time taken during constant speed and add the three times together to find the end result.

1. We know acceleration is 2 m/s², initial velocity 0, final velocity 50/3 m/s. We want to find both time and distance.
v = u + at will give us time:
50/3 = 0 + 2t
t = 25/3 s

v² = u² + 2ax will give us distance covered:
(50/3)² = 0² + 4x
x = 625/9 m

2. Note that the time taken to decelerate and the distance covered in this time is the same as in 1. You can try find these values manually again but they will turn out the same. (u = 50/3, v = 0, a = - 2)

3. So the distance covered during acceleration and deceleration = 625/9 + 625/9 = 1250/9 m. This leaves 1000 - 1250/9 = 7750/9 m in which the bus maintains a constant speed.

4. The time taken during which the bus maintains a constant speed is given by t = d/s = (7750/9) / (50/3) = 155/3 s.
Now we can find the total time for the whole trip: t(total) = 25/3 + 155/3 + 25/3 = 205/3 s = 68.3 seconds.

[allstar]

thanks phenomenol

I know this isn't a kinematics question, but I don't know how to change the subject title....
but...

what does kg wt mean?