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GCSE Physics notes: Explaining acceleration - graphs and calculations

2. Acceleration, deceleration, velocity-time graph interpretation, calculations and problem solving

Doc Brown's Physics Revision Notes

Suitable for GCSE/IGCSE Physics/Science courses or their equivalent

This page will help you answer questions e.g.  How do you define acceleration?

 How do you interpret velocity-time graphs?  How to do acceleration formula questions?   What to we mean by terminal velocity?

Sub-index for this page

(a) Brief reminders from section 1. on distance-time graphs

(b) Introducing velocity-time graphs and acceleration

(c) Example questions based on graph interpretation (not mathematical calculations)

(d) Acceleration formula, calculations, interpreting velocity-graphs (a = Δv Δt)

(e) More advanced calculations involving uniform acceleration (using v2 - u2 = 2as)

(f) Forces and circular motion - including acceleration and centripetal force



(a) Brief reminders from section 1. on distance-time graphs

Distance - time graphs

Graph shows acceleration, speeding up

Graph shows zero speed/velocity

Graph deceleration, slowing down

Graph shows uniform or constant speed/velocity, not acceleration

DO NOT confuse these graphs with the velocity-time graphs below!


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(b) Introducing velocity-time graphs and acceleration 

(DO NOT confuse these graphs with the ones above!)

Acceleration is about change in speed or velocity of an object and is not the same as speed or velocity.

The interpretation of the seven graphs above, by considering the gradient - positive, negative or zero,

and their mathematical sign if the values of acceleration (+) or deceleration (-) are used in calculations.

acceleration is the rate of change of velocity with time

acceleration = change in velocity time taken to effect the velocity change

a (m/s2) = Δv (m/s) Δt (s)    (actual numerical calculations are dealt with further down the page)

 

A comparison of speed/velocity - time graphs

Graph curves upwards, velocity is increasing (+ve), but also shows increasing acceleration, speeding up at an increasing rate.

 

A comparison of speed/velocity - time graphs

Graph is flat, shows constant velocity/speed (+ve), zero acceleration. Cruise control operating in a car.

 

A comparison of speed/velocity - time graphs

Graph shows decreasing acceleration (+ve), BUT not slowing down!, still speeding up but at a decreasing rate.

This is observed as you pull away in a car up to a maximum speed dictated by the speed limit.

 

A comparison of speed/velocity - time graphs

Graph is linear, shows constant or uniform acceleration (+ve), constant rate of increasing velocity.

 

A comparison of speed/velocity - time graphs

Graph is linear, shows constant or uniform deceleration (-ve), velocity decreasing, constant rate of deceleration.

 

A comparison of speed/velocity - time graphs

Graph shows decreasing deceleration (-ve), velocity decreasing, but decelerating at an increasingly slower rate.

 

A comparison of speed/velocity - time graphs

Graph shows increasing deceleration (-ve), velocity decreasing, but slowing down at an increasingly faster rate!

You can demonstrate this by gradually increasing the pressure on the brake pedal of a car and bringing it to sudden halt.

 

The seven graphs above illustrate of what you might see at any point on a velocity-time graph, BUT, in reality, any speed/velocity-time graph will be far more complicated than any one of these simple graphs. So, you may find a graph with all seven types of gradient in just one speed/velocity-time graph, like the velocity-time graph below which could represent a car journey.


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(c) Example questions based on graph interpretation (not involving mathematical calculations)

Q1.1 The graph below describes the velocity profile of a car journey.

Analyse the graph sections indicated below and describe the motion of the car at the various stages of the journey:

a-b: increasing acceleration,

b-c: linear/uniform/constant acceleration,

c-d: decreasing acceleration,

d-e: constant velocity, zero/no acceleration or deceleration

e-f: increasing deceleration

f-g: linear/uniform/constant deceleration,

g-h: decreasing deceleration

 

Q1.2 The sketch above illustrates the design of a roller coaster ride.

Using the numbers (1) to (9) and comment on energy store situations or changes ...

(a) Give two places where acceleration is taking place.

(4) and (5) accelerating downhill, conversion of gravitational potential energy (GPE) to kinetic energy (KE)

(b) At what points does the car have its maximum GPE?

at (3) and (7), the highest points you have the maximum in GPE

(c) At which point are you getting the greatest increase in GPE?

(1) => (2) => (3), the largest climb of the circuit.

(d) At which points do you get deceleration?

Climbing up at (6) as KE store decreases and GPE store increases.

(e) At which point do you get the maximum change of KE to increase a thermal energy store?

The braking zone <==(9)

 

Q1.3

 


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(d) Acceleration formula, problem solving calculations and interpreting velocity-graphs

Acceleration is about change in speed or velocity of an object and is not the same as speed or velocity.

Acceleration has its own defined formula, which expresses a change in velocity in a defined time interval.

acceleration = change in velocity time taken to effect the velocity change

a (m/s2) = Δv (m/s) Δt (s)

a = acceleration in metres per second squared m/s2Δv = change in speed/velocity in metres/second m/s

Δt = the time taken for the speed/velocity change in seconds, s

The formula for acceleration can also be written as:

a = (v - u) t = Δv t

where v = final speed/velocity,  u = initial speed/velocity,  t = time taken to change from velocity u to v

Rearrangements of acceleration formula: If a = Δv t   then   Δv = a x t    and   t = Δv a

 

Velocity - time graphs - illustrating the motion of objects A to F

From the graphs you can deduce several things:

From the gradient you can obtain the acceleration or deceleration (negative acceleration).

From the area under the 'triangles' or 'rectangles' of the velocity - time graph you can deduce the distance travelled - the displacement equals the area under a velocity time graph.

In terms of units, the distance calculation is equivalent to time x (distance / time)

e.g. = time x speed = s x m/s = m

velocity - time graphs linear uniform acceleration gradient calculations gcse physics igcse international courses

Graph (1) Linear or uniform acceleration

Both graph lines represent a linear increase in velocity AND the acceleration is constant.

From the equation a (m/s2) = Δv (m/s) Δt (s)

For object A: acceleration = (50 - 0)  (25 - 0) = 2.0 m/s2

For object B: acceleration = (50 - 30)  (20 - 0) = 1.0 m/s2

Note that the steeper the gradient, the greater the acceleration, the greater the rate of 'speeding up'.

Examples of distance travelled calculations from graph (1)

Here you are only dealing with an area of a triangle.

For object A between 0 and 20 seconds the distance travelled = (20 x 40) / 2 = 400 m

Because you are working with the area of a triangle, don't forget to halve the calculation!

For object B between 0 and 20 seconds the distance travelled = {20 x (50 - 30)} / 2 = 200 m

 

velocity - time graphs constant velocity zero acceleration gradient calculations gcse physics igcse international courses

Graph (2) Constant velocity - zero acceleration

The graph lines are horizontal, indicating there is no change in velocity, therefore zero acceleration.

Graph line C for object C represents a constant velocity of 20 m/s.

Graph line D represents an object D moving at twice the speed of C at 40 m/s.

Examples of distance travelled calculations from graph (2)

You are only dealing with a rectangle.

e.g. distance travelled by object C between 5 and 15 seconds = (15 - 5) x (20-0) = 200 m

and distance travelled by object D between 0 and 25 seconds = (25 - 0) x (40 - 0) = 1000 m

 

velocity - time graphs linear uniform negative acceleration deceleration gradient calculations gcse physics igcse international courses

Graph (3) Deceleration - negative acceleration

Both graph lines represent a linear decrease in velocity AND the deceleration (negative acceleration) is constant.

From the equation a (m/s2) = Δv (m/s) Δt (s)

For object E: acceleration = (0 - 40)  (25 - 0) = -1.6 m/s2

For object F: acceleration = (30 - 50)  (25 - 0) = -0.8 m/s2

Remember, if you express the answers as 1.6 m/s2 and 0.8 m/s2, it might be a good idea to say 'deceleration' or 'slowing down' too!

Note that the steeper the gradient, the greater the deceleration, the greater the rate of 'slowing down'.

Examples of distance travelled calculations from graph (3)

Again, like graph (1), it only involves the area of a triangle.

Distance travelled by object E between 0 and 25 seconds = 25 x (40 - 0) / 2 = 500 m

Distance travelled by object F between 10 and 25 seconds = (25 - 10) x (42 - 30) / 2 = 90 m

 

Note on use of words:

You can think of deceleration is negative acceleration, so when a moving object is slowing down the rate of change in speed/velocity with time is negative.

Therefore you can think of acceleration as the positive rate of change of speed/velocity.

But, you may find that the deceleration isn't used, and a negative sign used with the word acceleration.

So watch out for mathematical signs and how the words acceleration and deceleration are used.

 

Example calculations based on the acceleration formula and velocity-time graph interpretation

Q2.1 A car accelerates from 10 m/s to 30 m/s in 15 seconds.

Calculate the acceleration of the car.

a (m/s2) = Δv (m/s) Δt (s) a

If u and v are the initial and final velocities

a = (v - u) / t = (30 - 10) / 15 = 20 / 15 = 1.3 m/s2 (2 s.f.)

 

Q2.2 A train accelerates at 0.50 m/s2 for 30 seconds.

What is the increase in speed of the train?

a = Δv t   so   Δv = a x t

increase in speed = 0.5 x 30 = 15 m/s (2 s.f.)

 

Q2.3 If a car accelerates at 0.30 m/s2 from a standing start, how long will it take to attain a speed of 21 m/s?

a = Δv Δt   so   Δt = Δv a = (v - u) a, u and v are the initial and final velocities

time taken = (21 - 0) 0.30 = 21 s  (2 s.f.)

 

Q2.4 A car brakes sharply from moving at 30 m/s to 10 m/s in 4.0 seconds.

Calculate the deceleration of he car.

a = Δv Δt

If u and v are the initial and final velocities

a = (10 - 30) / 4.0 = 20 / 4 = -5.0 m/s2  (2 s.f.)

Note the minus sign for the acceleration - because the car is slowing down!

You can then say the deceleration is 5.0 m/s2  (2 s.f.)

 

Q2.5 The graph below summarises the first 100 seconds of a cyclist's race.

When calculating displacement (distance travelled), with this velocity - time graph, both triangles and rectangles are involved.

(a) (i) What is the initial uniform acceleration of the cyclist in this time period?

The initial acceleration is from 0 to 40 seconds.

speed = total change in speed / time taken = (10 - 0) / (40 - 0) = 10 / 40 = 0.25 m/s2 (2 s.f.)

(ii) How far has the cyclist travelled in this period?

From a speed/velocity-time graph you calculate distance covered from the area under the graph for the time period involved.

Area under the graph from 0 to 40 seconds = 10 x 40 / 2 = 200 m  (its easy if the graph is linear)

(b) At which point is the cyclist's acceleration the greatest? and calculate the acceleration at this point.

The steeper the gradients the greater the acceleration.

The gradient is steepest from 90 to 100 seconds, so that is the time period of greatest acceleration.

a = ∆v / ∆t = (25 - 20) / (100 - 90) = 5 / 10 = 0.5 m/s2

(c)

 

Q2.6 The graph below summarises a 100 minute commuter train journey, but with some signal delays!

Watch out for min ==> hour conversions!

(a) At what time or times is the train moving at constant speed? Explain your reasoning.

From the 20th to 40th minute AND from the 50th to 70th minute. The graph is horizontal, indicating constant speed (of 100 km/hour and 200 km/hour).

(b) Where is the greatest acceleration and what is its value in km/hour2?

Steepest gradient is from the 40th to the 50th minute.

Change in velocity = 200 - 100 = 100 km/hour

Time involved = 50 - 40 = 10 mins, 10/60 = 0.167 hours

gradient = 100 / 0.167 = 100 / 0.167 = ~600 km/hour2

(c) How far does the train travel in the first 20 minutes?

time = 20 / 60 = 0.333 hours, speed change = 100 - 0 = 100 km/hour.

area under graph = 100 x 0.333 / 2 = 16.7 km, ~17 km

 

Q2.7  A car travelling at 36 m/s skids off the road and hits a wall.

(a) If the car is brought to a halt in 2.0 seconds, what is the average deceleration of the car?

final velocity v = 0, initial velocity u = 36 m/s

acceleration a = (v - u) t = Δv t = (0 - 36) / 2.0 = -18 m/s2

So the average deceleration is 18 m/s2

Note: The acceleration due to gravity is 9.8 m/s2, so in this crash you body would experience a force of nearly '2G', not good for your body!

(b)

 

Q2.8 The graph below profiles the speed of a cyclist at the start of a race.

NOT a linear graph!

You get the distance travelled by calculating or measuring off the graph, the area under the graph for the specified time interval. To help you appreciate the method, I've highlighted two areas in yellow to be estimated to answer the two questions below.

Step 1 is to work out the distance relating to one small square.

From the 'bold' graph lines larger square you have 5 x 5 = 25 small squares = 2 m/s x 20 s = 40 m,

so each small square = 40 / 25 = 0.625 m, so from this you can estimate the areas and distances.

(a) How far does the cyclist travel in the first 20 seconds of the race? (give your answers to 2 s.f.)

0-10s ~ 2 squares, 10-20s ~9, total ~11 squares = ~11 x 0.625 = 6.88, distance travelled ~6.9 m

(b) How far does the cyclist travel between the 30th and 40th second of the race.

total small squares  = 25 + 22 (25-~3) + 5.5 = ~52.5, distance travelled = 52.5 x 0.625 = ~32.8 = ~32 m

Do you agree with my estimations?

 

Q2.9 A van is travelling at 20 m/s has its brakes applied for 5.0 seconds.

If this produces a deceleration of -1.5 m/s2, calculate the velocity of the van after braking.

a = Δv Δt

Δv = (v - u) = a x t  = -1.5 x 5.0 = -7.5

v - u = -7.5,  v - 20 = -7.5,  v = -7.5 + 20 = 12.5 m/s

 

Q2.10 If an object is fired vertically up in the air from ground level,

(a) Describe the energy store changes taking place and any assumptions you are making in your arguments. Assume the object leaves from, and returns to, the ground level.

When the object is thrown upwards it is given an initial maximum kinetic energy store.

As it rises the velocity decreases, its kinetic energy store (KE) decreases and its gravitational potential energy store (GPE) increases proportionately - this assume no other energy transfers are occurring.

This ignores the friction effect of air resistance - some KE is lost as thermal energy to increase the thermal energy store of the surroundings.

The decrease in velocity is due to the decelerating effect of the Earth's gravitational field.

When it reaches its maximum height, the object has its maximum GPE.

The object then accelerates back down to the ground, again due to the Earth's gravitational field.

At the same time, as the velocity increases, the object's GPE store is converted to its KE store and it hits the ground at its maximum velocity and with its maximum KE. This assumes no air resistance.

(b) Sketch the graph of velocity versus time for the whole flight.

There are two graph formats you can draw to illustrate the change in velocity with time and both are illustrated below.

velocity - time graph for throwing an object up in the air gcse physics igcse international gcse physics

Graph A is constructed in two halves.

From the maximum velocity x, the left half represents the decrease in velocity as the object's motion is slowed down by the Earth's gravitational field.

You can think of this as a negative acceleration i.e. deceleration - a negative gradient (a - m/s2).

The right half represents the increase in velocity as the object falls under gravity, attaining the maximum velocity x on hitting the ground.

You can think of this as a positive acceleration - a positive gradient (a + m/s2).

Graph A ignores the mathematical convention of a minus velocity on change of direction.

The bottom of the V, at time t, represents zero velocity as the object's motion is on the point of reversing direction to fall down.

Graph B is constructed as one continuous line.

This uses both positive and negative values of velocity, to represent the change in direction.

Therefore the velocity changes from +x to -x in a time of 2 x t, t being the point of zero velocity

 (c) Can you suggest a value for the gradients of the graph?

The gradients in each case would have the value of the Earth's gravitational acceleration/deceleration constant of 9.8 m/s2.

 

Q2.11


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(e) More advanced calculations involving uniform acceleration using v2 - u2 = 2as

(this section might not be required for your course)

Constant acceleration is known as linear uniform acceleration.

A good example is an object free falling in a gravitational field e.g. it is the acceleration due to gravity is 9.8 m/s2 at the Earth's surface.

You can use the following formula to do calculations based on a uniform acceleration situation.

An equation for uniform acceleration is ...

(final speed (m/s))2 - (initial speed (m/s))2 = 2 x acceleration (m/s2) x distance (m)

v2 - u2 = 2as, v = final velocity, u = initial velocity, a = acceleration, s = distance travelled or displaced

Rearrangements to calculate the various object variables in the above equation:

(units above, and remember in algebraic equations,

if you 'change the side you change the sign, or you change all the signs of the expression)

(i) v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

(ii) v2 - u2 = 2as,    -u2 = 2as - v2,    u2 = -2as + v2,    u = √(v2 - 2as)

(iii) v2 - u2 = 2as,   2as = v2 - u2,    a = (v2 - u2) 2s

(iv) v2 - u2 = 2as,   2as = v2 - u2,    s = (v2 - u2) 2a

 

Examples questions based on uniform acceleration

Q3.1 An initially stationary car accelerates uniformly at a rate of 1.50 m/s2.

(a) What is the speed of the car after a distance travelled of 100 m?

v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

Since u = 0 (initially car is stationary) and s = 100 m

v = √(2 x 1.5 x 100) = √300 = 17.3 m/s  (3 s.f.)

(b) If at 100 m the driver presses harder on the accelerator pedal to increase the acceleration to 2.00 m/s, what speed will the car be doing after travelling a total of 300 m from the start?

v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

u = 17.3 m/s (from part (a)), s = 300 - 100 (you must subtract 100 m because that's where new acceleration started

v = √{(2 x  2.0 x 200) + (17.32)} = √(800 + 300) = √1100 =  33.1 m/s  (3 s.f.)

Note the use of ( ) AND { } brackets, you really must take care about what follows the overall square root sign .

 

Q3.2 A jet rocket plane is launched from a converted jet airliner with an acceleration of 10.0 m/s2.

From radar tracking it is found that the rocket plane achieved a speed of 300 m/s after travelling for 1875 m.

At what speed was the launch aircraft travelling?

v2 - u2 = 2as,    -u2 = 2as - v2,    u2 = -2as + v2,    u = √(v2 - 2as)

Initially both the launch aircraft and the rocket plane are travelling at the same speed u.

v = 300 m/s, a = 10 m/s2, s = 1875 m

 u = √(v2 - 2as) = √{(3002) - (2 x 10.0 x 1875)}

u = √(90000 - 37500) = √52500 = 229 m/s  (3 s.f.)

 

Q3.3 A car is moving at a speed of 20 m/s when the driver presses on the accelerator pedal and after a distance of 400 m the car is travelling at 30 m/s. Assuming it is uniform, calculate the acceleration of the car.

v2 - u2 = 2as,   2as = v2 - u2,    a = (v2 - u2) 2s

s = 400 m, u = 20 m/s, v = 30 m/s (u and v being the initial and final velocities)

a = (302 - 202) / (2 x 400) = (900 - 400) / 800 = 0.625 = 0.63 m/s2  (2 s.f.)

 

Q3.4 A rocket is moving vertically upwards and leaves the launch silo with a velocity of 10.0 m/s. Booster rockets are immediately fired to produce an acceleration of 4.00 m/s2.

At what height (s) above the launch pad, will the rocket attain a velocity of 200 m/s (give your final answer in km).

(iv) v2 - u2 = 2as,   2as = v2 - u2,    s = (v2 - u2) 2a

a 4.00 m/s2, u = 10.0 m/s, v = 200 m/s  (u and v being the initial and final velocities)

s = (2002 - 102) / (2 x 4) = (40000 - 100) / 8 = 39900 / 8 = 4987.5 m = 4990 m or 4.99 km  (3 s.f.)

Note: To fire an object vertically from a 'powerful gun', to completely escape the Earth's gravitational field, requires an 'escape velocity' of 11 km/s at the end of the 'gun barrel'! However, rockets don't function like a gun. They have continuous thrust force from the rocket engines which will sustain a steady upward movement against the force of gravity.

 

Q3.5 An object is dropped from a height of 10 m above the ground. Ignoring air resistance, if the acceleration due to gravity is 9.8 m/s2, at what theoretical velocity does the object hit the ground?

v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

u = 0, v = ?, a = 9.8 m/s2, s = 10 m

since u = 0, v = √(2 x 9.8 x 10) = √196 = 14 m/s  (2 s.f.)

Note: (i) Not a good idea to fall when rock climbing, falling from ~31 ft and hitting the ground at 14 m/s. This will do a lot of damage, but no problem if all safety devices in place! BUT, imagine a 100 m sprinter running into a brick wall at an average speed of ~10 m/s!

(ii) The general formula for calculating this 'fall' velocity, since u is zero, is:

v = √(2as + u2) = √(2as) = √(2 x 9.8 x height) = √(19.6 x height above ground)

e.g. if you fall off a ladder at height of just 2 m above the ground (~6 ft) you hit the ground at:

v = √(2as) = √(2 x 9.8 x 2) = √39.2 = 6.3 m/s.

Comparing this with the speed of a sprinter, its hardly surprising the serious injuries you can suffer from a relatively short fall.

 

Q3.6 A an object is dropped from a certain height above the ground.

If the object hits the ground at 10 m/s, from what height was it dropped?

v2 - u2 = 2as, rearranging  s = (v2 - u2) 2a

s = height, v = 10 m/s, u = 0 m/s, a = 9.8 m/s2 (gravitational acceleration near Earth's surface)

height = s = (102 - 02)    (2 x 9.8) = 100   19.6 = 5.1 m

 

Q3.7 An object is fired vertically up into the air with an initial velocity of 20.0 m/s.

Ignoring air resistance, and taking the acceleration as -9.80/ms2, calculate the theoretical maximum height the object attains.

v2 - u2 = 2as, rearranging  s = (v2 - u2) 2a

s = (02 - 202)  (2 x -9.8)

s = -400  -19.6 = 20.4 m  (to 3 sf)

 

Q3.8 A falling object is travelling at 10 m/s when it is 10 m above the ground.

If gravitational acceleration is 9.8 m/s2, at what velocity will it hit the ground?

v2 - u2 = 2as,     v2 = 2as + u2,    v = √(2as + u2)

Substituting in the rearranged equation:

v = √{(2 x 9.8 x 10) + 102)

v = √(196 + 100) = √296

final velocity on hitting the ground = 17 m/s (2 sf)

 

Q3.9


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(f) Forces and circular motion - including acceleration and centripetal force

This section was adapted, re-edited and extended from the web page including a section on Gravity and circular motion.

Velocity is a vector quantity, it has both size (the speed) and direction.

If either the speed or direction changes, you have a change in velocity - you have an acceleration!

With this in mind, imagine whirling a conker around on the end of a piece of string, the moon orbiting Earth or planets orbiting the Sun.

What velocity are we dealing with? What force are we dealing with?

When an object goes round in a circle at a constant speed its direction is continually changing.

Even though the speed is constant, because direction changes, the velocity (a vector) is continually changing.

Therefore, if the velocity is changing, you must be dealing with an acceleration, even though the speed is constant.

To maintain this accelerating circular motion, there must be a force operating to maintain the circular path.

This is called the centripetal force - it can be the tension in a string as you whirl an object around or gravity holding some large object in orbit.

The direction of acceleration is inwards, the same direction as the centripetal force, and at 90o to the direction of motion.

(You don't have to know this, but the acceleration of an object moving at a constant speed in a circle is v2/r)

circukar motion and centripetal force changing velocity gcse physics igcse circular motion - velocity & centripetal force

To keep a body moving in a circle there must be a force directing it towards the centre.

This is called the centripetal force and produces the continuous change in direction of circular motion.

Even though the speed may be constant, the object is constantly accelerating because the direction is constantly changing via the circular path - i.e. the velocity is constantly changing (purple arrows, on the diagram).

For an object to be accelerated, it must be subjected to a force that can act on it - Newton's 1st law of motion.

Here the resultant centripetal force is acting towards the centre, so always directing the object to 'fall' towards the centre of motion (blue arrows on the diagram).

But the object is already moving, so the force causes it to change direction.

SO, the actual circular path of motion is determined by the resultant centripetal force (black arrows and circle) and the circling object keeps accelerating towards what it is orbiting.

The centripetal force stops the object from going off at a tangent in a straight line.

swinging an object round on a string circukar motion centripetal force changing velocity gcse physics igcse Swinging something round on a string.

When you swing something round on the end of a string, the tension in the string is the centripetal force.

Imagine whirling a conker round on the end of a string.

You yourself feel this force of tension as the 'pull' in the string (I've marked in a black line to represent the string).

If you could use a fast action camera to monitor the motion and the string broke, you would observe the object would fly off at the precise tangent to the circular path and in a straight line of constant velocity - the result resultant of Newton's 1st law!

Since gravity and air friction act on the object, you do have to keep on 'inputting' kinetic energy to keep it swinging round.

The centripetal force will vary with the mass of the object, the speed of the object and the radius of the path the object takes.

 

The same arguments on circular motion apply to the movements of planets around a sun, a moon around a planet and a satellite orbiting a planet. The orbits are usually elliptical, rarely a perfect circle, but the physics is the same.

In these cases, it is the force of gravitational attraction that provides the centripetal force and it acts at right angles to the direction of motion.

You should also realise that they are moving through empty space (vacuum), so there are no forces of friction to slow the object down.

This is why the planets keep going around the Sun and the moon keeps going around the Earth.

When satellites are put into orbit they are given just the right amount of horizontal velocity so that the resultant centripetal force of gravity keeps the satellite in its a circular orbit.

You can vary this horizontal velocity to position satellites at different distances above the Earth's surface.


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Motion and associated forces notes index (including Newton's Laws of Motion)

1. Speed and velocity - the relationship between distance and time, distance-time graphs gcse physics

2. Acceleration, velocity-time graph interpretation and calculations, problem solving gcse physics revision notes

3. Acceleration, friction, drag effects and terminal velocity experiments gcse physics revision notes

4. Newton's First, Second and Third Laws of Motion, inertia and F = ma calculations gcse physics revision notes

5. Reaction times stopping distances and example calculations gcse physics revision notes

6. Elastic and non-elastic collisions, momentum calculations and Newton's 2nd law of motion gcse physics notes


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