Velocity, acceleration and displacement

This equation applies to objects in uniform acceleration:

This is when:

  • final velocity (v) is measured in metres per second (m/s)
  • initial velocity (u) is measured in metres per second (m/s)
  • acceleration (a) is measured in metres per second squared (m/s²)
  • displacement (x) is measured in metres (m)

Calculating final velocity

The equation above can be used to calculate the final velocity of an object if its initial velocity, acceleration and displacement are known. To do this, simplify the equation to find v:

  • v^{2} = u^{2} + 2\:a\: x
  • v = \sqrt{u^{2}+2ax}

Example

A biscuit is dropped 300 m, from rest, from the Eiffel tower. Calculate its final velocity, ignoring air resistance. (Acceleration due to gravity = 10 m/s².)

v^{2} = u^{2} + 2\:a\: x

v = \sqrt{u^{2}+2ax}

v = \sqrt{0^{2}+2 \times10} \times 300

v =\sqrt{6000}

\underline{v =77.5~m/s}

Calculating acceleration

The equation can also be used to calculate the acceleration of an object if its initial and final velocities, and the displacement are known. To do this, rearrange the equation to find a:

  • v^{2}-u^{2}=2\ a\ x
  • a=\frac{v^{2}-u^{2}}{2x}

Example

A train accelerates uniformly from rest to 24 m/s on a straight part of the track. It travels 1.44 km. Calculate its acceleration.

  1. v^{2}-u^{2}=2\ a\ x
  2. a=\frac{v^{2}-u^{2}}{2x}
  3. a = \frac{24^{2}-0^{2}}{2 \times 1440}
  4. a = 576 \div 2880
  5. \underline{a = 0.2\ m/s^{2}}

Calculating other quantities

The equation can also be rearranged to find initial velocity (u) and displacement (x):

  1. u = \sqrt{v^{2}-2ax}
  2. x = \frac{v^{2}-u^{2}}{2a}

These equations of motion:

  1. acceleration = \frac{change\ in\ velocity}{time\ taken}
  2. speed = \frac{distance\ travelled}{time}

can be rearranged to form other useful equations to help work out.

If an object is changing speed then this equation

distance\ travelled = speed \times time

becomes

distance\ travelled = average\ speed \times time

distance\ travelled = \frac{1}{2} (initial\ speed + final\ speed) \times time

or

x = \frac{1}{2}(u + v) \times t

This equation

acceleration = \frac{change\ in\ velocity}{time\ taken}

can be written:

a = \frac{(v-u)}{t}

and rearranged as:

v=u+at

Higher - Combining these two equations gives

x = ut+\frac{1}{2}at^{2}

Question

A toy rocket is fired vertically upwards against gravity with an initial velocity of 20 m/s. What height will it reach in 1 second if a = -10 m/s²? Ignore air resistance in your calculation.

x = ut+\frac{1}{2}at^{2}

ut = 20 \times 1 = 20

\frac{1}{2}\ at^{2} = 0.5 \times- 10 \times (1)^{2} = \text{-}5

x = 20\text{-}5=15~m