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Equations of motion for SQA National 5 Physics
This page covers the following topics:
1. Basic equation of motion
2. Projectile motion
For an object moving at a constant velocity, the distance it has travelled during a given time interval at a given speed can be calculated using the following formula: distance = speed × time. The instantaneous velocity of an object is its velocity at a specific point in time during its motion, whereas its average velocity is the average of all instantaneous velocities during the object's motion.
An object thrown into the air follows a parabolic path called a trajectory and is called a projectile. This object will move under the influence of gravity.
Since gravity acts downwards, it only affects the vertical velocity of the object. Equations of motion can be used to calculate the vertical component of velocity. The object travels upwards at a decreasing rate, until its velocity becomes 0 and it starts to accelerate towards the ground until it reaches it.
There are no horizontal forces acting on the projectile, since air resistance is usually neglected, so the horizontal component of velocity remains constant during the time of flight and the equation velocity = distance ÷ time can be used.
1
How long must the motion of an object travelling at a constant speed of 14 m/s be to cover 65 m?
distance = speed × time
t = 65 ÷ 14 = 4.64 s (3 s. f.)
4.64 s
2
Define instantaneous velocity.
The instantaneous velocity of an object is its velocity at a specific point in time.
velocity at a specific time
3
An object covers 25 m in 8 seconds. Calculate the average speed of the object.
Using d = s × t,
average speed = 25 m/ 8 s = 3.125 m/s.
3.125 m/s
4
A projectile is moving in a parabolic path. Explain why its vertical velocity is changing, whereas its horizontal velocity remains constant throughout its flight.
Since the projectile is moving under the influence of gravity, it has a downwards acceleration of 9.8 m/s² throughout its flight. By Newton's second law, since there is a resultant force acting in the vertical direction, the vertical velocity will be changing. There are no horizontal forces acting on the object, therefore by Newton's first law, the horizontal velocity will remain constant.
vertical weight → Newton’s 2nd law → downwards acceleration, no horizontal friction → Newton’s 1st law → constant horizontal velocity
5
A stone is thrown off a cliff horizontally at a velocity of 8 m/s. Given that the stone lands into the ocean at a horizontal distance of 15 m from the cliff and that g = 9.8 m/s², find the height of the cliff.
horizontally: s = d ÷ t
8 m/s = 15 m ÷ t
time of flight = 15m ÷ 8 m/s = 1.875 s
vertically: s = at² ÷ 2
s = 9.8 m/s² × (1.875 s)² ÷ 2 = 17.2 m (3 s. f.)
17.2 m
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