Acceleration
accelerationThe rate of change in speed (or velocity) is measured in metres per second squared. Acceleration = change of velocity Ă· time taken. is the rate of change of velocity. It is the amount that velocity changes per unit time.
The change in velocity can be calculated using the equation:
change in velocity = final velocity - initial velocity
\(\Delta v = v - u\)
The average acceleration of an object can be calculated using the equation:
\(acceleration = \frac{change~in~velocity}{time~taken}\)
\(a = \frac{\Delta v}{time~taken}\)
This is when:
- acceleration (a) is measured in metres per second squared (m/sÂČ)
- change in velocity (âv) is measured in metres per second (m/s) (â is the Greek letter delta, representing 'change in')
- time taken (t) is measured in seconds (s)
If an object is slowing down, it is decelerationSlowing down or negative acceleration, eg the car slowed down with a deceleration of 2 msâ»ÂČ. (and its acceleration has a negative value).
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Example
A car takes 8.0 s to accelerate from rest to 28 m/s. Calculate the average acceleration of the car.
final velocity, v = 28 m/s
initial velocity, u = 0 m/s (because it was at rest - not moving)
change in velocity, âv = (28 â 0) = 28 m/s
\(a = \frac{\Delta v}{t}\)
\(a = \frac{28}{8}\)
\(a = 3.5~m/s^2\)
Question
A car takes 25 s to accelerate from 20 m/s to 30 m/s. Calculate the acceleration of the car.
final velocity, \(v = 30~m/s\)
initial velocity, \(u = 20~m/s\)
change in velocity, \(\Delta v = (30 â 20) = 10~m/s\)
\(a = \frac{\Delta v}{t}\)
\(a = \frac{10}{25}\)
\(a = 0.4~m/s^2\)
Velocity, acceleration and distance
This equation applies to objects in uniform acceleration:
(final velocity2) - (initial velocity2) = 2 Ă acceleration Ă distance
\(v^2 - u^2 = 2~a~s\)
This is when:
- final velocityThe speed, in a particular direction, of a body after it accelerates (after it changes speed or direction). (v) is measured in metres per second (m/s)
- initial velocityThe speed, in a particular direction, of a body before it accelerates (before it changes speed or direction). (u) is measured in metres per second (m/s)
- acceleration (a) is measured in metres per second squared (m/s2)
- distance (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:
\(v^2 = u^2 + 2~a~s\)
\(v = \sqrt{u^2 + 2as}\)
Example
A biscuit is dropped 320 m, from rest, from the Eiffel tower. Calculate its final velocity. (Acceleration due to gravity = 10 m/s2.)
\(v^2 = u^2 + 2~a~s\)
\(v = \sqrt{u^2 + 2as}\)
\(v = \sqrt{0^2 + 2 \times 10 \times 320 }\)
\(v = \sqrt{6400}\)
\(v = 80~m/s\)
Calculating acceleration
The equation can also be used to calculate the acceleration of an object if its initial and final velocities, and the distance 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 Ă 1,000 = 1,440 m
- Then use the values in the equation:
\(v^2 = u^2 + 2~a~s\)
\(a = \frac{v^2 - u^2}{2s}\)
\(a = \frac{24^2 - 0^2}{2 \times 1,440}\)
\(a = \frac{576}{2,880}\)
\(a = 0.2~m/s^2\)
A car manufacturer claims that a car can accelerate from 0 km/h to 100 km/h in 4.0 s.
Calculate its acceleration.
- first convert km/h to m/s: 100 km/h = 100 Ă· 3.6 = 27.8 m/s
- then use the values in the equation:
\(a = \frac{\Delta v}{t}\)
\(a = \frac{27.8 - 0}{4.0}\)
\(a = 6.95~m/s^2\)
Calculating other quantities
The equation can also be rearranged to find initial velocity (u) and distance (s):
\(u^2 = v^2 - 2~a~s\)
\(s = \frac{v^2 - u^2}{2a}\)