Velocity, acceleration and distance

This equation applies to objects in uniform acceleration:

()² – ()² = 2 × acceleration × distance

\(v^{2} – u^{2} = 2αx\)

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 (α) 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, rearrange the equation to find v:

\(v^{2} – u^{2} = 2 α x\)

\(v = \sqrt{ u^{2} +{2α x} } \)

Example

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

\(v^{2} – u^{2} = 2 α x\)

\(v = \sqrt{ u^{2} +{2α x} }\)

\(v = \sqrt{ {0}^{2} + {2} \times {10} \times 300 }\)

\(v = \sqrt{6,000}\)

\(v = {77.5} \ m/s\)

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 α:

\(v^{2} – u^{2} = 2 α x\)

\(α = \frac{v^{2} – u^{2}} {2x}\)

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. First convert km to m:

1.44 km = 1.44 × 1,000 = 1,440 m

2.Then use the values in the equation:

\(v^{2} – u^{2} = 2 α x\)

\(α = \frac{v^{2} – u^{2}} {2x}\)

\(α = \frac{24^{2} – 0^{2}} {2 \times 1,440}\)

\(α = 576 ÷ 2,880\)

\(α = 0.2 m/s^{2}\)

Calculating other quantities

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

\(u = \sqrt{ v^{2} - {2α x} } \)

\(x = \frac{v^{2} – u^{2}} {2α}\)

��˿��

Motion - EdexcelVelocity, acceleration and distance