Rotational Motion & Astrophysics: Kinematic Relationships

Question 1

What is displacement (with respect to time), and what can it be defined as?

Answer 1

The displacement (s) of the particle is a function of time: s = f(t)

(Meaning it moves with respect to time)

The displacement (s) is the length and direction of a particle from its origin.

Question 2

What is velocity (with respect to displacement), and how can average vs. instantaneous velocity be found?

Answer 2

Velocity is the rate of change of displacement.

Average velocity can therefore be found by equation: v = △s/△t

Instantaneous velocity is calculated by deriving the displacement: v = ds/dt

Question 3

What is acceleration (with respect to velocity), and how can average vs. instantaneous acceleration be found?

Answer 3

Acceleration is the rate of change of velocity

Average acceleration can therefore be found by equation: a = △v/△t

Instantaneous acceleration is calculated by deriving the velocity: a = dv/dt

Question 4

Derive the equation of motion v = u + at

use a = d²s/dt² as a starting point

Answer 4

Integrate with respect to time:

∫ (d²s/dt²)dt = ∫ (a)dt

To get:

ds/dt = at + k

At t = 0: ds/dt = k, so k = u

At t = t: ds/dt = v

Thus, v = u + at

Question 5

Derive the equation of motion s = ut + 1/2at²

use the first equation of motion as a starting point

Answer 5

Integrate with respect to time:

∫ (ds/dt)dt = ∫ u + at dt

s = ut + 1/2at² + k

at t = 0: s = 0 therefore k = 0

Thus, s = ut + 1/2at²

Question 6

Derive the equation of motion v² = u² + 2as

use the first equation of motion as a starting point

Answer 6

Square both sides:

v² = (u+at)² = (u+at)(u+at)

v² = u² + 2uat + a²t²

v² = u² + 2a(ut + 1/2at²)

substitute the brackets for the second equation of motion, s

Thus, v = u² + 2as

Question 7

What can we find by deriving the gradient of a displacement-time graph?

(The area under the graph has no meaning)

Answer 7

The gradient of a displacement-time graph gives us the instantaneous velocity

(Calculated by v = ds/dt)

Question 8

What can we find by deriving the gradient of a velocity-time graph?

Furthermore, what can we find by integration of such a graph between limits?

Answer 8

The gradient of a velocity-time graph gives us the instantaneous acceleration

(Calculated by a = dv/dt)

Integrating between limits gives us the displacement (between limits)

(Calculated by ∫ v dt)

Question 9

What can we find by integration of an acceleration-time graph between limits?

(The gradient has no meaning)

Answer 9

Integrating between limits gives us the change in velocity (between limits)

(Calculated by ∫ a dt)