Equations of motion

The equations of motion relate to the following five quantities:

  • u – initial velocity
  • v – final velocity
  • a – acceleration
  • t – time
  • s – displacement

Of the above u, v, a and s are vector quantities. As such, remember to make vectors going in one direction positive and vectors in the opposite direction negative.

Time (t) is a scalar quantity.

If any three of the five quantities are known, then the other two may be calculated using the following four equations:

  • v = u + at
  • x = ut + ½at2
  • x = (u+v) ÷ 2 × t
  • v2 = u2 + 2as

The equations of motion can only be used for an object travelling in a straight line with a constant acceleration.

Question

A nail is fired from a nail gun into a fixed block of wood. The nail has a speed of 380 ms-1 just as it enters the wood.

The nail comes to rest after penetrating 60 mm into the wood.

Find the time taken for the nail to come to rest. Assume that the retarding force on the nail is constant as it penetrates the wood.

Vector diagram where left to right is the positive direction for vectors. U equals 380 metres per second left to right. V equals zero. S equals 60 millimetres. A and t are both unknown.

u = 380 ms-1

v = 0

a = ?

t = ?

s = 60 × 10-3

v2 = u2 + 2as

02 = 3802 + 2a × 60 × 10-3

a = -1,203,333 ms-2

v = u +at

0 = 380 + (-1,203,333)t

t = 0.0003157

The time taken for the nail to come to rest is 3.16 × 10-4 s.

Question

A ball is projected vertically upwards with an initial velocity of 24.5 ms-1. The effects of air resistance may be neglected.

Vector diagram where upwards is positive direction for vectors. U equals 24.5 metres per second upwards. V equals zero. A equals 9.8 metres per second per second downwards. T and s are both unknown.

Calculate the time taken for the ball to reach its maximum height.

Let unknown vector quantities go in the positive direction for vectors. Vertical motion is constant acceleration of 9.8 ms-2 towards the centre of the Earth.

u = 24.5 ms-1

v = 0

a = - 9.8 ms-2

t = ?

s = ?

v = u + at

0 = 24.5 + (-9.8)t

t = 2.5 s

Question

Now calculate the maximum height reached by the ball.

u = 24.5 ms-1

v = 0

a = -9.8 ms-2

t = 2.5 s

s = ?

s = ut + ½at2

s = 24.5 × 2.5 + ½(-9.8) 2.52

s = 30.625

Maximum height reached is 31 m.