大象传媒

How can we describe motion? - OCR 21st CenturyAcceleration

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.

Part of Physics (Single Science)Explaining motion

Acceleration

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/s虏)
  • change in velocity (鈭唙) is measured in metres per second (m/s) (鈭 is the Greek letter delta, representing 'change in')
  • time taken (t) is measured in seconds (s)

If an object is slowing down, it is (and its acceleration has a negative value).

Learn more on acceleration in this podcast

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, 鈭唙 = (28 鈥 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.

Velocity, acceleration and distance

This equation applies to objects in uniform acceleration:

(final velocity2) - (initial velocity2) = 2 脳 acceleration 脳 distance

\(v^2 - u^2 = 2~a~s\)

This is when:

  • (v) is measured in metres per second (m/s)
  • (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.

  1. First convert km to m: 1.44 km = 1.44 脳 1,000 = 1,440 m
  2. 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.

  1. first convert km/h to m/s: 100 km/h = 100 梅 3.6 = 27.8 m/s
  2. 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):

\(u^2 = v^2 - 2~a~s\)

\(s = \frac{v^2 - u^2}{2a}\)