Rates of change

Rates of change are questions involving realworld concepts such as distance (or height), speed, velocity, acceleration and deceleration. The best way to start questions involving rates of change is to find all three general formulas for distance, speed and acceleration. The distance (sometimes height) formula is typically given to you in the question. There are a few things that you need to learn for this section: Example:

An object is fired vertically upwards and its height, h metres, after any number of seconds, t , is given by the formula h = 48t – 2t2. Find:

(i) the height of the object after 3 seconds

(ii) its speed after 3 seconds

(iii)its acceleration

(iv) after how many seconds is the object momentarily at rest. (v) the maximum height reached by the object. Solution:

(i) Height = 48t –2 t2

Sub in t =3:

Height = 48(3) – 2(3)

= 144 – 2(9) = 126m

(ii) Speed = 48 – 4t Speed = 48 – 4(3) Speed = 48 – 12 = 36 m/s

(iii)Acceleration = – 4 m/s2

Here the acceleration is actually deceleration since the answer is negative.

(iv) At rest means that Speed = 0 ⇒

=0 48–4 t =0

–4 t =–48 t = 12 seconds

(v) Max height occurs when speed = 0 t = 12 seconds (from part (iv)) height when t = 12 seconds Height = 48t –2 t2

= 48(12) – 2(12)2

= 576 – 2(144)

= 576 – 288 = 288m

Don’t forget about the units of measurement. These are easy marks to lose.