# Equations of uniformly accelerated motion, with worked examples

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

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 (
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 :) .