#suvat

[gatheryepens]

Suvat equations rearranged

[xxx-cards-xxx]

[rafisics]

SUVAT Calculator

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[physicsandchemistryrevision]

I don't get kinematics at all . A level physics sucks :( I keep failing its so depressing

Kinematics is difficult, yes. I still struggle with parts of it to this day if I’m honest! But don’t lose faith – there is still hope!

I find,the main trick to it is identifying what you have and what you need to know. Particularly with SUVAT-related questions, you will want to try to draw from the question distance s; initial and final veocity, u and v respectively; acceleration a and time t. Often you will be given information such as, “the car brakes until it is stationary,” meaning you know that v = 0 m/s because it’s final velocity is when it is stationary. You will start to pick up on these things the more questions you do.

Irecently posted a general guide to solving kinematics problems and a worked example of a standard kinematics question tohelp you:

Kinematics help

Kinematics example

and see a kinematics question I recently answered:

Kinematics along an incline

Luckily,I have a few older posts on this blog which could be of some help! Unfortunately, I don’thave many worked examples.

G481 motion/kinematics

Work and power (uses SUVAT)

Redefining SUVAT

Instantaneous velocity and acceleration

Vectors:

Vector calculations

Vector quantites and Scalar quantites

Trignometry(useful for vector resolution):

Vector triangles

Resultant vectors (i.e. resolving vectors)

Triangle theorems

I hopethis helps! If you want me to walk you through some kinematics examples you’restuck on, let me know and I’ll answer the best I can!

[physicsandchemistryrevision]

Mathematics

Kinematics example

See "G481 Mechanics" and "Previous worked questions"

This post is best viewed on my blog, click here to be taken to its permalink.

(a) A ball is launched with initial velocity v0 at an angle θ to the horizontal and lands on a shelf at a height yshelf from the plane of the launch point r0. Assume air resistance effects are negligible.

Find an expression for the total horizontal distance travelled, xmax.

The first thing to notice is the wording of the question: we only have to determine an expression (i.e. an equation) which describes xmax – no numerical values required at this stage.

Let’s picture the scenario.

Now we can set this to a Cartesian (xy) plane. For simplicity, place the centre of mass of the ball of the initial launch point at the origin of the axes.

Now, a key part of kinematics is treating the flight path as a regular function. Doing this we can analyse it in the same way as we would a function. Hence, we can infer that the turning point is at dy/dx = 0 which allows us to find the maximum height of the flight, for example.

To find the points where the curve is equal to the height of the shelf we find the points where the function intersects y = yshelf.

Clearly we can see that f(x) intersects the line yshelf at two discrete (i.e. separate and unique) points in x. Now, how do we find these points?

This leads us to another fundamental and very useful property of kinematics: Time is spatially independent. This means that time doesn’t depend on x or y exclusively so when x = x1 at time t1, y = f(x1) at t1. Hence, time is used as a way of ‘linking’ the x and y components. If this doesn’t yet make sense, you’ll see what I mean in a little while.

So, recall the SUVAT equations. I won’t write them here to save space but click on the link if you’re unsure or interested. What we want to do is find the times at which the function f(x) (or, the ball) crosses the line y = yshelf and find what the value of x is at those times. One of the x values will be xmax and the other will be x1.

Okay, so we want to know time t. We know position y at this time and we know the initial velocity v0. Hence, we select the SUVAT equation

r = v0 t + ½ a t2

because it is dependent on position r, initial velocity, acceleration a (which is simply the acceleration due to gravity, – g, since we have an absence of external or resistive forces) and, what we want to find, time t.

Since r = (x, y) and we know y but not x, we shall select the y-component of the above equation,

y = v0, yt – ½ gy t2.

Note that since g only acts in the y-direction, we’ll simply let gy = g.

Alternatively, we could express this as a function f(x), as shown on the above diagram

f(x) = v0, yt – ½ g t2

and find the point where our function is equal to yshelf

yshelf = f(xmax)

yshelf = v0, yt – ½ g t2.

With rearrangment, we can use the quadratic roots formula to find the values of t where the two functions intersect.

0 = – ½ g t2 + v0, yt – yshelf

Hence, solving for t gives us,

t = [– v0, y ± √(v0, y2 – 4 (– ½ g) (–yshelf)] / [2 (– ½ g)]

t = [v0, y ∓ √(v0, y2 – 2 g yshelf)] / g.

We can select each case of t by selecting + √(v0, y2 – 2 g yshelf) and – √(v0, y2 – 2 g yshelf) separately:

t1 = [v0, y – √(v0, y2 – 2 g yshelf)] / g,

t2 = [v0, y + √(v0, y2 – 2 g yshelf)] / g.

Now, we can analyse these to select the time for (xmax, yshelf), which will be the one of greatest value, t2. Hence, our time tmax to reach xmax is given by

tmax = [v0, y + √(v0, y2 – 2 g yshelf)] / g.

We should now look back at the SUVAT equations and select one which contains position r and time t, ideally (but not essentially) disregarding velocity v. Therefore, we shall use

r = v0 t + ½ a t2

and by selecting the x component, we get the resolved equation

x = v0, xt – ½ gx t2.

Since we know that gx = 0 and, at x = xmax, t = tmax, our equation becomes

xmax = v0, xtmax – ½ (0) tmax2

xmax = v0, xtmax.

Using our previously determined expression for tmax we find an expression for xmax,

xmax = v0, x [v0, y + √(v0, y2 – 2 g yshelf)] / g,

where, using resolution of vectors, we can express v0, x and v0, y explicity,

v0, x = |v0| cos θ, v0, y = |v0| sin θ.

Therefore,

xmax = {v0 cos θ [v0 sin θ + √(v02 sin θ – 2 g yshelf)]} / g,

where |v0| = v0.

xmax = [v02 sin θ cos θ + v0 cos θ √(v02 sin θ – 2 g yshelf)] / g,

                                                   _______________ xmax = [½ v02 sin 2θ + v0 cos θ √v02 sin θ – 2 g yshelf] / g.

Although assumptions and approximations could be made to simplify the equation, this suffices as an expression for xmax and is only dependent on v0, θ and yshelf. If we were considering potential energy and kinetic energy exchanges, we could simplify this expression to an extent.

(b) The ball was fired with it’s magnitude of the initial velocity v0 = 2 m/s at an angle of 40° from the horizontal. The shelf is at a height of 2 metres.

(i) Find an expression for distance to the shelf from r0 = (0, 0).

The distance travelled is given by the magnitude of position, |r| or r.

Hence, by Pythagoras’ theorem, we know that

r2 = x2 + y2

and so

r = (xmax2 + yshelf2)1/2.

By substituting xmax we get,

r = ( [¼ v04 sin2 2θ + v0­2 cos2θ (v02 sin θ – 2 g yshelf)] / [g 2] + yshelf2 )1/2.

(ii) Find a value for r.

r = ( {¼ (2)4 sin2 2(40) + (2)­2 cos2 (40) [(2)2 sin (40) – 2 (9.81) (2)]} / {(9.81) 2} + (2)2 )1/2.

r = 2.22 m

(iii) Find a value for xmax.

We could either rearrange the entire above formula, or we can use our value for r and rearrange the Pythagorean theorem. For simplicity, we’ll do the latter.

xmax = (yshelf2 – r2)1/2,

xmax = ( 22 – 1.222 )1/2,

xmax = 1.58 m.

[physicsandchemistryrevision]

Physics

Redefining SUVAT Cont'd from "Instantaneous velocity and acceleration" 

Some of the equations we're deriving may seem familiar to you but with a new 'twist'. We'll start by redefining some of the quantities from my post about instantaneous velocity and acceleration and making some important assumptions.

x2 = x, x1 = x0

t2 = t, t1 = 0

v2 = v, v1 = v0

a = constant, so a = Δv/Δt = (v - v0)/(t - 0)

and by rearranging a, we get

① v = v0 + at 

(we know this as the first SUVAT equation)

We can define average velocity vav  in two ways: in terms of v and v0 and in terms of Δx and Δt. Hence, we can equate these

vav = ½(v + v0 ) = Δx/Δt

Taking the ½(v + v0 ), we can substitute equation ① for v

½(v + v0 ) = ½(v0 + at + v0 ) = ½(2v0 + at) = v0 + ½at

and equate as before

v0 + ½at = Δx/Δt

v0 + ½at = (x - x0)/t

by rearranging, we can get an expression for x in terms of a, t, v0 and x0

② x = x0 + v0t + ½at2

(we know this as the second SUVAT equation with an new term of x0)

We'll rearrange equation ② 

x - x0 = v0t + ½at2

and our expression for a for t

t = (v - v0)/a

and substitute t into our rearranged expression of ② 

x - x0 = v0(v - v0)/a + ½a(v - v0)²/a²

x - x0 = 2v0(v - v0)/2a + (v - v0)²/2a

2a(x - x0 ) = 2v0(v - v0) + (v - v0)²

2a(x - x0 ) = 2v0v - 2v0² + v² - 2v0v + v0²

and so, by canceling 2v0v and one v0², we get 

③ 2a(x - x0 ) = v² - v0²

(we know this as the third SUVAT equation with an new term of x0)

An alternative way of reaching equation ① is be defining the velocity in terms of the integral of instantaneous acceleration with limits t, 0, v and v0.

{a = dv/dt } · dt

{a dt = dv } · ∫t=0t=t

∫0t a dt' = ∫v0vdv' 

[at ]0t = [v ]v

a(t - 0) = v - v0

① v = v0 + at

and by following the same technique for v, we can reach a new equation.

{v = dx/dt } · dt

{v dt = dx } · ∫t=0t=t

∫0t v dt' = ∫x0xdx' 

∫0t (v0 + at) dt = ∫0t v0 dt + ∫0t at dt

[v0t + ½at²]0t = [x ]x0x

v0t + ½at² = x - x0

so, by rearranging

④ x = v0t + ½at² + x0

(we know this as the fourth SUVAT equation with an new term of x0)

[physicsandchemistryrevision]

AS Maths M1

Quick notes SUVAT equations

v² = u² + 2as

v = u + at

s = ½(u + v)t

s = ut + ½at²

s = vt - ½at²

where s is displacement (interchangeable with x, y, h, and r in 2 dimensions), u is initial velocity, v in final velocity, a is acceleration and t is time.

[wolfantian]

i know ill try to distract myself with tumblr so where best to leave my revision?