2. Kinematics

Question 1

Once again, only quick notes D:

Time is ticking fr

Answer 1

D:

:(

Question 2

Define velocity?

Answer 2

The change of displacement per unit time

Question 3

How to find average speed/velocity?

Answer 3

S = D/T

And that’s why it’s m/s O_o

Question 4

Define acceleration?

Answer 4

The change of velocity per unit time

Question 5

How to find average acceleration?

Answer 5

m/s² = m/s/s

that’s why it’s m/s² O_o

Question 6

Know about displacement time graph

Answer 6

1. Gradient gives velocity (remember finding average speed)
2. Displacement based on distance away from origin, negative side = reversing type shi
3. Constant = z
   Stationary = line
   Acceleration = arc upwards
   Deceleration = parabowler

Question 7

Know about velocity time graph

Answer 7

1. Gradient gives acceleration (remember finding average acceleration)
2. However area gives the displacement sooo, don’t get confused
3. Constant acceleration = z
   Stationary = constant speed
   Deceleration = z going down
   Ig no arcs?

Question 8

How to derive the suvat equations?
(simplified first)

Answer 8

SUVAT
(S = vt - 1/2 at²) not involved (no u))
^^ lol that’s just there for display

• They’re in the data booklet
• And u kinda just add random stuff together and see what u get
• But other than that it’s kinda just, if u know u know

Question 9

How to derive V = u + at (without s)

Answer 9

V = u + at (without s)
- Gradient = rise/run (from velocity time graph):
1. a = (v - u)/t
- v = final velocity (the end)
- u = initial velocity (at the beginning)
- a = just acceleration cuz velocity/time
- t = time, part of velocity/time
2. Re-arrange for v = u + at
- Self explanatory, look in data booklet for full formula

Question 10

How to derive S = 1/2 (u+v) t (without a)

Answer 10

S = 1/2 (u+v) t (without a)
- Displacement = average velocity x time
- Area of trapezium x time (use previous suvat equation a = (v-u)/t)
1. S = (u + v) x t/2
2. U can simple change the /2 into a half, the real equation
- S = 1/2 (u+v) x t

Question 11

How to derive S = ut + 1/2 at² (without v)

Answer 11

S = ut + 1/2 at² (without v)
- Combine S = 1/2 (u+v) t & V = u + at
(Subbing in)
1. S = 1/2 (u + {u + at}) x t
2. S = 1/2 (U + U + at) x t
3. S = 1/2 (2u + at) x t
4. S = 1/2 (u + at) x t
5. S = ut + 1/2 at²
- Typical re-arranging to get the final equation

Question 12

How to derive V² = u² + 2as (without t)

Answer 12

V² = u² + 2as (without t)
- Use S = 1/2 (u+v) t & V = u + at
- However change v = u + at
- For finding t = (v-u)/a
- Then just sub that in
1. S = 1/2 (u + v) x (v - u)/a
2. S = (u +v)x(v-u)/2a
- All divided by 2a
3. 2as = uv + v² - u² - uv
4. 2as = v² - u²
5. u² + 2as = v²
6. V² = u² + 2as
- As long as u know where to start first
- Then the rest u can just re-arrange by just looking at the final equation
- Unsure if this gives a lotta marks, I’d be pleased

Question 13

We have a problem, projectile motion. I’d wanna ask for help for this one but ahhh can I at least write summin brief notes?

Answer 13

• Say no air resistance or summin
• Sometimes asked to find horizontal & vertical components (z sin/cos theta)
• Use suvat for height or time of flight
• Range = total horizontal distance
  use d = st
• Down arc = parabowler
• for acceleration (9.81), going up is negative, going down is positive

Question 14

What is projectile motion?

Answer 14

• Projectile = any body that is given an initial velocity
• and then follows a path determined by the effect of gravity on that object
• Path followed by that object = trajectory

Question 15

What are the 2 motions consisting within projectile motion?

Answer 15

• Horizontal motion
• Vertical motion

Question 16

Tell me about horizontal motion in projectile motion

Answer 16

• Not affected by gravity
• Remains constant
  1. Initial velocity = ucosθ
  2. Acceleration = 0
  3. Displacement at time t
• Use S = ut + 1/2 at²
• We know a = 0
• And to find horizontal displacement = uxt
• ∴ x = (ucosθ)t
  4. Final velocity = same as initial velocity
• Using v = u + at
• a = 0
• v_x = u_x = ucosθ
  dude just use ucosθ for horizontal component

Question 17

Tell me about vertical motion in projectile motion

Answer 17

• Subject to deceleration going up
• Acceleration going down due to gravity
  1. Initial velocity = usinθ
  2. Acceleration = -g (initially & -9.81)
  3. Displacement at time t
• Use S = ut + 1/2 at²
• We know a = -g
• Vertical displacement would be y = u_yt - 1/2 gt²
• ∴ y = (usinθ)t - 1/2 gt²
  4. Final velocity = initial minus gt
• Using v = u + at
• a = -g
• v_y = u_x
  dude just use usinθ for vertical component

Question 18

How to analyze projectile motion?

Answer 18

Analyzing questions:
- So in general, we’re given either:
1. A normal trajectory type thing
2. Trajectory from a cliff (parabowler)
3. Normal trajectory but elevated
- In addition, remember this can only apply without air resistance
- And apparently it’s better to already set out suvat symbols already

Question 19

How to answer typical projectile motion question?

Answer 19

So for typical questions:
1:
1. Calculate magnitude of horizontal and vertical components
- Horizontal = ucosθ
- Vertical = usinθ
- Remember both different things for certain questions
2. Calculate maximum height
- Only look at vertical side, not horizontal
- g = -9.81
- v = 0 (final velocity)
- Re-arrange equation v² = u² + 2as for S
- S = (v² - u²)/2a
all divided by 2
- In the end should give maximum height
3. Calculate time of flight
Just to make sure, times ur answer by 2 since we only get one half of the time
- Vertical side only still
- Same values as to before
- Use equation v = u + at and rearrange for t
- (v - u)/a = t
- Afterwards, times answer by 2
4. Calculate range of cannon ball
- Now this ones for the horizontal side only
Which would only consist of: s, u, t
- We already found the initial velocity
- Time we take from finding time of flight
- Use s = d/t and rearrange for d
d = st
- Voila!

Question 20

How to answer projectile motion from cliff?

Answer 20

2 (From cliff):
1. Shape = parabowler
2. Time taken for shell to hit water
- Gain horizontal and vertical components always just incase
- Only look at vertical side for this
- g = 9.81 (we’re going down)
- s is given from question hopefully
- Use s = ut + 1/2 at²
Works because initial is always 0 for cliff trajectory
- And just rearrange for t
3. Calculate range of shell
- Same as previous ones, we got the t from vertical and we already worked out horizontal component
- s = d/t -> d = st
4. Calculate magnitude & direction of resultant velocity
- This is actually them vector triangles type shi
- Base is already horizontal component
- So we need to find the height, how?:
Find velocity from equation v = u + at
(use data from vertical motion stuff we did earlier)
- After gaining the “height” (just velocity), do pythagoras theorem to find the magnitude
a² + b² = c²
- And then u can gain the direction using Soh Cah Toa

Question 21

How to answer projectile motion of an elevated trajectory?

Answer 21

3 (Elevated trajectory):
1. Basketball, fadeaway shot, how long it takes for ball to hit ground?
- So this is just purely vertical motion part
- Find vertical component
- Afterwards, hope that it states how high it was released
- Then apparently that height has gotta be negative (could be similar to how going up must be -g)
- Use equation S = ut + 1/2 at²
- And then, u actually just re-arrange to make it into ax² + bx + c
- Gain values of t by factorising (or use calculator B|)
- Whatever answer u get, obviously it won’t be the negative
(Don’t get confused, keep trying to figure it out, took me a while from the exemplar question)