# Suvat questions and their purpose

Could someone help me interpret the information derived from the suvat questions.

I understand where they’re derived from but i don’t quite understand what their purpose is. My textbook doesn’t help offer much insight.

Regards
Original post by KingRich
Could someone help me interpret the information derived from the suvat questions.

I understand where they’re derived from but i don’t quite understand what their purpose is. My textbook doesn’t help offer much insight.

Regards

Not sure what you mean. The simplest example of suvat is vertical motion under gravity, so motion when force/acceleration is constant which means the velocity is linear and the displacement is a quadratic function of time. So suvat motion is a simple/important special case of more general variable force/acceleration motion.

Its also the simplest generalization of the speed-distance-time (so zero acceleration) motion.
(edited 1 year ago)
Original post by KingRich
Could someone help me interpret the information derived from the suvat questions.

I understand where they’re derived from but i don’t quite understand what their purpose is. My textbook doesn’t help offer much insight.

Regards

Do you mean 'what is the purpose of SUVAT equations?' ??

SUVAT equations are 'shortcuts' to relating displacement, velocity, acceleration, and time of an object together in the strict case where acceleration is constant (uniform) for that object.

Remembering them helps you answer mechanics related questions in the context of motion where forces are not directly considered (this is called kinematics).

E.g. a problem may be posed whereby (very often) you have to identify what is given to you (s, u, v, a, or t) and decide what you need to determine, and then think about which SUVAT equation(s) help you do that, and then use them carefully within the context.

One common context is projectile motion because objects under the influence of gravity are treated to have a constant acceleration of approx. 9.8 m/s^2 towards the ground, so you can expect to use SUVAT a lot in these types of contexts.

I don't know what stage of your A-Level you are at, but in Year 2 you learn that in the case of non-constant acceleration we see that displacement, velocity, and acceleration are all related via the more powerful and general tool of calculus ... and it just so happens that if you treat acceleration as a constant in this scope then SUVAT equations fall out of calculus as a consequence.
(edited 1 year ago)
Original post by RDKGames
Do you mean 'what is the purpose of SUVAT equations?' ??

SUVAT equations are 'shortcuts' to relating displacement, velocity, acceleration, and time of an object together in the strict case where acceleration is constant (uniform) for that object.

Remembering them helps you answer mechanics related questions in the context of motion where forces are not directly considered (this is called kinematics).

E.g. a problem may be posed whereby (very often) you have to identify what is given to you (s, u, v, a, or t) and decide what you need to determine, and then think about which SUVAT equation(s) help you do that, and then use them carefully within the context.

One common context is projectile motion because objects under the influence of gravity are treated to have a constant acceleration of approx. 9.8 m/s^2 towards the ground, so you can expect to use SUVAT a lot in these types of contexts.

I don't know what stage of your A-Level you are at, but in Year 2 you learn that in the case of non-constant acceleration we see that displacement, velocity, and acceleration are all related via the more powerful and general tool of calculus ... and it just so happens that if you treat acceleration as a constant in this scope then SUVAT equations fall out of calculus as a consequence.

Yes, this is what I was looking for. Thank you for explaining.

I sorta skipped the mechanic section in book 1 to continue pure maths in book 2 but I’ve decided to revisit book 1 mechanics to mix it up. I’ve only covered kinematics so far.

so, for these specific suvat questions, they only work when a particle is constant. Could you give me an example for me to picture? Would throwing a stone of a cliff be a good example of this as I’d expect the acceleration to be constant due to obviously not having a way to increase its acceleration(I.e: engine)

Although, if I’m not mistaken, I think objects increase in velocity when falling from a great height. Eh. I’ll await your response lol
Original post by KingRich
Yes, this is what I was looking for. Thank you for explaining.

I sorta skipped the mechanic section in book 1 to continue pure maths in book 2 but I’ve decided to revisit book 1 mechanics to mix it up. I’ve only covered kinematics so far.

so, for these specific suvat questions, they only work when a particle is constant. Could you give me an example for me to picture? Would throwing a stone of a cliff be a good example of this as I’d expect the acceleration to be constant due to obviously not having a way to increase its acceleration(I.e: engine)

Although, if I’m not mistaken, I think objects increase in velocity when falling from a great height. Eh. I’ll await your response lol

The bit in bold doesn't actually make sense! SUVAT equations are valid when acceleration is constant.

Broadly speaking, acceleration is change in velocity (speed) over a time period, so for constant acceleration we can write a = (v - u)/t where u is initial velocity (at the start of the time period) and v is final velocity (i.e. at the end of the time period). This can be immediately rearranged to give v = u + at, which is the simplest of the SUVAT equations.

The other equations can be derived by considering the definition of distance travelled as area under a graph, or as a simple exercise in basic calculus where a = dv/dt, v = ds/dt etc and imposing some initial conditions e.g. distance travelled = 0 at time t = 0.
Original post by KingRich
Yes, this is what I was looking for. Thank you for explaining.

I sorta skipped the mechanic section in book 1 to continue pure maths in book 2 but I’ve decided to revisit book 1 mechanics to mix it up. I’ve only covered kinematics so far.

so, for these specific suvat questions, they only work when a particle is constant. Could you give me an example for me to picture? Would throwing a stone of a cliff be a good example of this as I’d expect the acceleration to be constant due to obviously not having a way to increase its acceleration(I.e: engine)

Although, if I’m not mistaken, I think objects increase in velocity when falling from a great height. Eh. I’ll await your response lol

* Displacement says where a particle is when measuring from some fixed point.

* Velocity says how much displacement changes per unit time. It also tells us in which direction the particule is moving.

* Acceleration says how much velocity changes per unit time.

SUVAT questions must contain constant acceleration, so yes the velocity can be changing; no problem here. Velocity is increasing if acceleration is positive, or decreasing when acceleration is negative. Velocity is constant when acceleration is 0.

Yeah a stone being thrown off a cliff is a good example of SUVAT, however for this you need to model the problem in 2D because the particle can move horizontally AND vertically in this context.

A simpler example: consider dropping a stone from rest into an empty well and hearing a contact sound 10 seconds later, then using this to work out how deep the well is.

It's a 1D problem because the motion is purely vertical.

Here you know u=0, t=10, and since it's a projectile it will have a = -9.8 due to gravity ... and what you're after is s (displacement of the stone from the top of the well to the point of contact, hence the depth of the well).

For this you need to recall that s has a few formulae to choose from:

s = ut + 1/2 at^2
s = vt - 1/2 at^2
s = 1/2 (u + v)t

and the only one fitting our information is the first one. Subbing in u,t,a gives you a displacement of -490 ... which means the well is 490 metres deep.

It is important in SUVAT questions to choose which direction is positive and which one is negative. In my example here I do what I always do and choose UPWARDS to be positive direction, which is why my acceleration is negative since it is acting DOWNWARDS. Likewise, the displacement is negative since it is measured from the top of the well DOWN to the bottom.

You can very well choose downwards to be the positive direction and hence choose a = 9.8 and get a positive displacement of 490 instead. In either case, we deduce that the well is 490m deep.
(edited 1 year ago)