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Acceleration for SQA Higher Physics
This page covers the following topics:
1. Newton’s second law
2. Interpreting velocity-time graphs
3. Gradient of velocity-time graphs
Newton's second law states that the acceleration of an object is directly proportional to the resultant force in its direction and is inversely proportional to the mass of the object. It can be expressed using the following formula: resultant force = mass × acceleration. This equation can only be used when the acceleration of the moving object is constant.
A velocity-time graph is a graphical representation of an object's motion over a given period of time. The direction and the speed of an object can be deduced from the velocity-time graph of its motion. The area under the velocity-time graph of an object's motion gives its displacement.
The gradient of a velocity-time graph gives the acceleration of an object. A positive gradient shows that the object is accelerating, whereas a negative one shows that it is decelerating.
1
The velocity-time graph of an object's motion is drawn as a horizontal line. Explain what this says about the object's motion.
The object is travelling at constant velocity.
constant velocity
2
An object is travelling at a constant velocity of 15 m/s for the first 6 seconds of its motion and then it slows down at a constant rate to rest in a further 5 seconds. Draw a velocity-time graph for this object.
The y axis of the graph is velocity in m/s and the x axis is time in seconds.
The object travels at a 15 m/s velocity for 6 seconds.
The object then decelerates to rest in a further 5 seconds.
image
3
Plot a labelled velocity-time graph for an object travelling at a constant rate of 17 m/s for 2 seconds and then accelerates for 5 seconds at a rate of 1 m/s² and travels at this new velocity for a further 3 seconds.
The y axis of the graph is velocity in m/s and the x axis is time in seconds.
The object travels at a 17 m/s velocity for 2 seconds.
The object then accelerates to 17 m/s + 5 × 1m/s² = 22 m/s in 5 seconds.
The object continues travelling at a constant 22 m/s velocity for another 3 seconds.
image
4
Calculate the acceleration of an object of mass 7 kg which has a resultant force of 28 N acting on it .
Using F = ma, 28 N = 7 kg × a, thus acceleration = 28 N/7 kg = 4 m/s².
4 m/s²
5
A car of mass 1800 kg is travelling across a road with a forward thrust force of 2000 N. The road exerts a frictional force of 200 N. Calculate the acceleration of the car.
resultant force = 2000 N − 200 N = 1800 N
Using F = ma, 1800 N = 1800 kg × a.
acceleration = 1800 N/1800 kg = 1 m/s²
1 m/s²
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