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###### mechanics

Original post by TrES2b

does achieving two answers always mean that the particle travelled through the point twice. Basically what I'm asking is that if I find two answers for t would this always be the meaning?

Yes. Your displacement - time suvat equation is a quadratic, so you should expect two answers if you're solving for t.

Original post by TrES2b

also with this problem how do i know that the negative accerlation means its going backwards and not that its slowing down?

Negative acceleration will mean that the velocity becomes increasingly negative. If the current velocity is positive, then this does mean that it is slowing down. If the current velocity is negative, then it becomes larger negative. Slowing down generally refers to speed rather than velocity, so just be careful with terminology.

(edited 1 year ago)

Original post by mqb2766

Negative acceleration will mean that the velocity becomes increasingly negative. If the current velocity is positive, then this does mean that it is slowing down. If the current velocity is negative, then it becomes larger negative. Slows down generally refers to speed rather than velocity, so just be careful.

im still confused on how the two answers for t are correct and how I'm supposed to know it returned back to the starting point?

Original post by TrES2b

im still confused on how the two answers for t are correct and how I'm supposed to know it returned back to the starting point?

You solved for s=0 which is the starting point (zero displacement at t=0 is the trivial solution).

(edited 1 year ago)

Original post by mqb2766

You solved for s=0 which is the starting point (zero displacement at t=0 is the trivial solution).

for part a for example 9 s=20 but I got two answers for t?

Original post by TrES2b

for part a for example 9 s=20 but I got two answers for t?

Didnt know which part you were talking about, but yes there are two solutions again. Initially the velocity is positive and passes through A heading right (first value of t). The acceleration is negative, so the velocity becomes zero at the maximum (positive) displacement and then returns to A with negative velocity heading left (second value of t).

(edited 1 year ago)

Original post by mqb2766

Didnt know which part you were talking about, but yes there are two solutions again. Initially the velocity is positive and passes through A heading right (first value of t). The acceleration is negative, so the velocity becomes zero at the maximum (positive) displacement and then returns to A with negative velocity heading left (second value of t).

but how will i know it returend to the starting point because in the question it didn't tell me that. like I'm confused does deceleration = slows doing or going backwards

Original post by TrES2b

but how will i know it returend to the starting point because in the question it didn't tell me that. like I'm confused does deceleration = slows doing or going backwards

The relationship between deceleration (negative acceleration) and velocity is given by the simple, linear suvat

v = u + at

If "a" is negative, then the velocity-time line has a negative gradient. If "u" is positive, then you keep decreasing it. v will get smaller ("slows down"), be zero at some point in time and then become increasingly negative ("speeds up" in the opposite direction to the initial positive motion). Every 1 unit of time, v is reduced by a fixed amount |a|.

As in the previous post, slowing down is generally a scalar concept, it does not refer to the direction and so usually refers to speed. However, suvat ... uses velocity which has both magnitude and direction, so you have to be a bit careful about interpreting it like that.

(edited 1 year ago)

Original post by mqb2766

The relationship between deceleration (negative acceleration) and velocity is given by the simple, linear suvat

v = u + at

If "a" is negative, then the line has a negative gradient. If "u" is positive, then you keep decreasing it. v will get smaller ("slows down"), be zero at some point in time and then become increasingly negative ("speeds up" in the opposite direction to the initial positive motion). Every 1 unit of time, v is reduced by a fixed amount |a|.

As in the previous post, slowing down is generally a scalar concept, it does not refer to the direction and so usually refers to speed. However, suvat ... uses velocity which has both magnitude and direction, so you have to be a bit careful about interpreting it like that.

v = u + at

If "a" is negative, then the line has a negative gradient. If "u" is positive, then you keep decreasing it. v will get smaller ("slows down"), be zero at some point in time and then become increasingly negative ("speeds up" in the opposite direction to the initial positive motion). Every 1 unit of time, v is reduced by a fixed amount |a|.

As in the previous post, slowing down is generally a scalar concept, it does not refer to the direction and so usually refers to speed. However, suvat ... uses velocity which has both magnitude and direction, so you have to be a bit careful about interpreting it like that.

ohhhhh thank youuu

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