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In the example of the express train at the start of this subsection, we have initial velocity u = 0 km/h, final velocity v = 300 km/h and time taken t = 300 s. So 300 − 0 acceleration a = = 1 km/h per second. Worked 300 example 2.6 uses the more standard velocity units of m/s.
Other units for acceleration are possible. Earlier we saw examples of acceleration in mph per second and km/h per second, but these are unconventional. It is usually best to work in m/s2.
Studyy tip p Units of acceleration
Acceleration is a vector quantity – it has a direction. It can be forwards (positive) or backwards (negative). So it is important always to think about velocity rather than speed when working out accelerations, because velocity is also a vector quantity.
In Worked example 2.6, the units of acceleration are given as m/s2 (metres per second squared). These are the standard units of acceleration. The calculation shows that the aircraft’s velocity increased by 2 m/s every second, or by 2 metres per second per second. It is simplest to write this as 2 m/s2, but you may prefer to think of it as 2 m/s per second, as this emphasises the meaning of acceleration.
Questions Worked example 2.6
2.13 Which of the following could not be a unit of acceleration? km/s2, mph/s, km/s, m/s2 2.14 A car sets off from traffic lights. It reaches a speed of 27 m/s in 18 s. What is its acceleration? 2.15 A train, initially moving at 12 m/s, speeds up to 36 m/s in 120 s. What is its acceleration?
An aircraft accelerates from 100 m/s to 300 m/s in 100 s. What is its acceleration? Step 1: Start by writing down what you know, and what you want to know. initial velocity u = 100 m/s final velocity v = 300 m/s time t = 100 s acceleration a = ? Step 2: Now calculate the change in velocity. change in velocity = 300 m/s − 100 m/s = 200 m/s Step 3: Substitute into the equation. change in velocity acceleration = time taken 200 m/s = 100 s = 2 m/s2 Alternatively, you could substitute the values of u, v and t directly into the equation. v u a= t 300 − 100 = = 2 m/s2 100
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Cambridge IGCSE Physics
Acceleration from speed–time graphs A speed–time graph with a steep slope shows that the speed is changing rapidly – the acceleration is greater. It follows that we can find the acceleration of an object by calculating the gradient of its speed–time graph: acceleration = gradient of speed – time graph Three points should be noted: The object must be travelling in a straight line; its velocity is changing but its direction is not. ◆ If the speed–time graph is curved (rather than a straight line), the acceleration is changing. ◆ If the graph is sloping down, the object is decelerating. The gradient of the graph is negative. So a deceleration is a negative acceleration. ◆
Original material © Cambridge University Press 2014
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