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# Kinematics

Equations of motion involving acceleration

Acceleration is a measure of the rate of increase in velocity of a body and the acceleration in this section will be considered to be uniform i.e. constant for a given set of circumstances.

acceleration = increase in velocity
time taken

Example 5: The velocity of a body increases from 2 m/s to 18 m/s in 8 s. Determine the acceleration of the body.

Increase in velocity = 18 - 2
= 16 m/s

acceleration = 16 m/s
8 s

acceleration = 2 m/s²

Note that the units of acceleration m/s² are obtained by dividing m/s by s.

Retardation

Bodies also slow down and this is called deceleration or retardation. Retardation is the term that is usually used. Retardation is the reverse of acceleration and therefore can be written as negative acceleration.

Example 1: The velocity of a body decreases from 25 m/s to 5 m/s in 4 s. Determine the retardation of the body.

Change in velocity = 5 - 25

= -20 m/s

Therefore acceleration = -20 m/s
4s

acceleration = -5m/s²
or   retardation = 5 m/s²

Various equations of motion can be derived as follows:

acceleration = increase in velocity/time taken
i.e.    a = (v - u)/t
or     at = v - u
or     v = u + at (i)

average velocity = (u + v)/2 because the acceleration is uniform.
Also average velocity = s/t

Therefore                   s/t = (u + v)/2
or                                s = (u+v)t/2 (ii)

Substitute for v = u + at in (ii) gives:

s = (u + u + at)t/2
or                                s = ut + ½at ² (iii)

Substitute for t = (v - u)/a in (ii) gives:
s = (u + v)/2 x (v - u)/a
or                 2as = (u + v)(v - u)
or                 2as = v² - u²
or                 v² = u² + 2as (iv)

The above four equations are used to solve problems involving uniform acceleration in kinematics.

Strictly speaking s represents displacement in the above formulae. In practice, s can often be used to denote distance because if the direction of motion does not change then the displacement and distance travelled are equal.

When solving problems in kinematics it is often useful if the given information is written down in terms of a, u, v, t and s. This then helps in the choice of which equation(s) to use for the solution.

Example 7: A car has an acceleration of 3m/s². If the initial velocity of the car is 5 m/s, determine:
(a) How far the car travels in 6 s;
(b) How far the car has travelled when it reaches a velocity of 30 m/s.

(a) Known Values
u = 5 m/s;   a = 3 m/s²;   t = 6 s;   s = Unknown

use   s = ut + ½ at²
so     s = 5 x 6 + ½ x 3 x 6²
or     s = 84 m

(b) Known Values
u = 5 m/s;   a = 3 m/s²;   v = 30 m/s;   s = Unknown

use   v² = u² + 2as
so    30² = 5² + 2 x 3 x s
or    s = 145.8 m