Describing motion - AQAVelocity, acceleration and distance
The movement of objects can be described using motion graphs and numerical values. These are both used to help in the design of faster and more efficient vehicles.
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 (s) 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:
\(v^2 = u^2 + 2~a~s\)
\(v = \sqrt{u^2 + 2as}\)
Examples
A biscuit is dropped 300 m, from rest, from the Eiffel tower. Calculate its final velocity. (Acceleration due to gravity = 9.8 m/s2.)
\(v^2 = u^2 + 2~a~s\)
\(v = \sqrt{u^2 + 2as}\)
\(v = \sqrt{0^2 + 2 \times 9.8 \times 300}\)
\(v = \sqrt{5880}\)
\(v = 76.7~m/s\)
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:
\(v^2 - u^2 = 2~a~s\)
\(a = \frac{v^2 - u^2}{2s}\)
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.
First convert km to m:
1.44 km = 1.44 脳 1,000 = 1,440 m
Then use the values in the equation:
\(v^2 - u^2 = 2~a~s\)
\(a = \frac{v^2 - u^2}{2s}\)
\(a = \frac{24^2 - 0^2}{2 \times 1,440}\)
\(a = 576 \div 2,880\)
\(a = 0.2~m/s^2\)
Calculating other quantities
The equation can also be rearranged to find initial velocity (u) and displacement (s):