NOTES p.14

Deriving the Equations of Motion

The following 3 equations can be used for any situation that involves a constant acceleration (horizontally OR vertically).

** MEMORIZE THE DERIVATIONS**

GRAPH FOR CONSTANT ACCELERATION (in general)

velocity

(ms-1)

time (s)

u

v

t

u = initial velocity (ms-1)

v = final velocity (ms-1)

a = acceleration (ms-2)

s = displacement (m)

t = time (s)

velocity

(m/s)

time (s)

u

v

t

1st Equation: “a” = rate of change of velocity

a = (v - u) / t

so v = u + at

velocity

(ms-1)

time (s)

u

v

t

2nd Equation: “s” = area under graph line

Area A

Area B

s = Area A + Area B

s = ut + ½ (v-u) t

s = ut + ½ (at) t

s = ut + ½ a t2

v2 = (u + at)2

v2 = u2 + 2uat + (at)2

v2 = u2 + 2uat + a2 t2

v2 = u2 + 2a (ut + ½ a t2)

v2 = u2 + 2as

velocity

(m/s)

time (s)

u

v

t

3rd Equation: SQUARE EQUATION 1

Area A

Area B

v2 = u2 + 2as

s = ut + ½ a t2

v = u + at For all calculations …

1st Decide a SIGN CONVENTION

2nd List “u v a s t”

3rd Select equation(s) and calculate

UP = + u = 2 ms-1

v = ?

a = - 9.8 ms-2

s = - 30 m

t = ?

Worked Example

A helicopter travelling upwards at 2 ms-1 releases a box from a height of 30 m. Calculate the landing velocity of the box.

v2 = u2 + 2as

= 22 +(2 x - 9.8 x -30)

= 592

v = ±24.3 ms-1

v = 24.3 ms-1 , down

v2 = u2 + 2as

Problems to do: 33 – 42 on Equations of Motion

s = ut + ½ a t2

v = u + at For all calculations …

1st Decide a SIGN CONVENTION

2nd List “u v a s t”

3rd Select equation(s) and calculate

UP = + or - u =

v =

a =

s =

t =

v2 = u2 + 2as

Problems to do: 33 – 42 on Equations of Motion

s = ut + ½ a t2

v = u + at33. 280m 34. 51.2

m/s

35. 28 m/s 36. 16.7 s

37. 3.03 s

38. a) 750 m/s2 b) 0.04 s

39. 9.5 N/kg

40. a) 0.21 m/s2 b) 1.43 s

41. 234 m

42. a) (i) 21.4 m (ii) 15.6 m/s

b) 34.6 m