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

Doc Brown's Physics Revision Notes

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

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?

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

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

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 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.

Example questions based on graph interpretation  (initially 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

Acceleration formula and calculations

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

Note that deceleration is negative acceleration, so when a moving object is slowing down the change in velocity is negative, so watch your mathematical signs in acceleration calculations. Therefore acceleration is the positive rate of change of speed/velocity.

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.

(a) 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.)

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 ths 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 ?

More advanced calculations involving speed and acceleration  (this section might not be required for your course)

Constant acceleration is known as 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

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

For acceleration and circular motion see

could repeat the section here?

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

gcse physics

gcse physics revision notes

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

gcse physics revision notes

gcse physics revision notes

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