**If this is your first time please review Speed, velocity and acceleration from scratch**

**First equation of motion**

Suppose a body which is already moving with a velocity of

*u*in m /s begins to accelerate

at the rate of

*a*in m / s ². The velocity will now increase by the numerical value of

*a*in m/s for

each second that it moves. The increase in velocity in a time

*t*in s will therefore be equal to

*at*.

Hence the final velocity at the end of t is given by

**v = u + at .............................................. ( 1 )**

This is called the

*first equation of motion.*

**Second equation of motion**

If a body is moving with uniform acceleration its average velocity is equal to half the sum

of the initial velocity

*u*, and the final velocity

*v.*

Thus,

**average velocity = (u + v )/ 2**

but v = u + at

therefore

**average velocity = ( u + u + at )/2 = u + 1/2 at**

**The distance, x, moved ( displacement )= average velocity × time**

x = u + 1/2 at × t

or

**x = u + 1/2 at² ................( 2 )**

This is known as the

*second equation of motion.*

**Third equation of motion**

A useful third equation can be obtained by eliminating

*t*between the first two equations.

Squaring both sides of the equation, v = u + at, we obtain

**v² = u² + 2 uat + a² t²**

Taking out the factor

*2a*from the last two terms of the right-hand side,

v² = u² + 2 a(ut + a t²/2)

but the bracket term is equal to x

hence,

**v² = u² + 2 ax...........................(3)**

Well, here they are :

v = u + at .............................................. ( 1 )

x = u + 1/2 at² ................( 2 )

v² = u² + 2 ax...........................(3)

**Worked examples**

1.

*A stone is thrown vertically upwards with an initial velocity of 14 m/s. Neglecting air resistance*,

**find**; (a) the maximum height

reached: (b) the time taken before it reaches the ground.

( Acceleration due to gravity = 9.8 m/s² .)

Solution

Solution

**Free tip**: When working problems of this type the reader is recommended to

**extract**the data in the question and write them down against the appropriate symbols

**before**attempting to substitute in one of the equations of motion.

*u = 14 m/s*

v= 0 m/s

a = - 9.8 m/s² (

v= 0 m/s

a = - 9.8 m/s² (

**retardation**

*)*

To find the

**height reached**, x, we substitute in the equation

*v² - u² = 2 ax*

0² - 14² = 2 × ( - 9.8 ) × x

whence

x = - 14²/ 2 × ( - 9.8 ) = 10 m.

The time taken to reach this height is found by substitution in the equation

*v = u + at*

Thus, 0 = 14 + ( - 9.8 ) × t

or t = 14/ 9.8 = 1.43 s

The time taken to fall back to the ground is found by substitution in

*x = ut + 1/2 at²*

in which x = 10 m

u = 0 m/s

a = + 9.8 m/s²

Thus,

10 = 0 × t + 1/2 × 9.8 × t²

or t² = 10 × 2 / 9.8

whence t = 1.43 s

the downward motion is, of course, simply a reversal of the upward motion in every respect.

2.

*A car starts from rest and is accelerated uniformly at the rate of 2 m/s² for 6 s. It then maintains a constant speed for half a minute. The brakes are then applied and the vehicle uniformly retarded to rest in 5 s.*

**Find**the maximum speed reached in km/h and the total distance covered in meters.

**Solution**

**First stage**

u = 0 m/s

a = 2 m/s²

t = 6 s

Substituting in the first equation of motion,

*v = u + at*

v = 0 + 2 × 6

v = 12 m /s

= 12 / 1000 × 60 × 60 km /h

= 43 km /h

The distance moved in the first stage may be found by substituting in the second equation of motion, thus,

*x = ut + 1 /2 a t²*

= 0 × 6 + 1/2 × 6² = 36 m

**Second stage**

u = 12 m/s ( constant )

t = 30 s

hence

*distance moved = speed × time*

= 12 × 30 = 360 m.

**Third stage**

u = 12 m/s

v = 0 m/s

t = 5 s

Acceleration a = ( v - u )/ t = ( 0 - 12) / 5 = - 2.4 m/s²

The distance may be found either by second or third equations of motion. If we use the latter,

*v² = u² + 2 ax*

whence

x = v² - u² / 2 a = 0² - 12² / 2 × ( -2.4) =

*30 m. I hope that was simple :) .*