2. Acceleration, velocitytime 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 velocitytime
graphs?
How to do acceleration formula
questions?
What to we mean by terminal velocity?
Brief reminders from section 1. on distancetime
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
velocitytime graphs below!
Velocitytime
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/s^{2}) = 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 velocitytime graph, BUT, in reality, any
speed/velocitytime 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/velocitytime graph, like the velocitytime 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:
ab: increasing acceleration,
bc: linear/uniform/constant acceleration,
cd: decreasing acceleration,
de: constant velocity, zero/no acceleration or
deceleration
ef: increasing deceleration
fg: linear/uniform/constant deceleration,
gh: 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/s^{2}) = Δv
(m/s) ÷ Δt (s)
a = acceleration in metres per second squared m/s^{2},
Δv = 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 velocitytime 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/s^{2}) = 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/s^{2}
(2 s.f.)
Q2.2 A
train accelerates at 0.50 m/s^{2} 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/s^{2} 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/s^{2}
(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/s^{2}
(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/s^{2}
(2 s.f.)
How far has the cyclist travelled in this period?
From a speed/velocitytime 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/s^{2}
(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/hour^{2}?
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/hour^{2}
(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/s^{2}
So the average deceleration is 18 m/s^{2}
Note: The acceleration due to gravity
is 9.8 m/s^{2}, 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.)
010s ~ 2 squares, 1020s ~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/s^{2}
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/s^{2}) x distance (m)
v^{2}  u^{2}
= 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) v^{2}  u^{2} = 2as,
v^{2} = 2as + u^{2},
v = √(2as + u^{2})
(ii) v^{2}  u^{2} = 2as,
u^{2} = 2as  v^{2}, u^{2} = 2as
+ v^{2},
u =
√(v^{2}  2as)
(iii) v^{2}  u^{2} =
2as, 2as = v^{2}  u^{2},
a = (v^{2}  u^{2})
÷ 2s
(iv) v^{2}  u^{2} = 2as,
2as = v^{2}  u^{2},
s
= (v^{2}  u^{2}) ÷ 2a
Q3.1 An
initially stationary car accelerates uniformly at a rate of 1.50 m/s^{2}.
(a) What is the speed of the car after a distance travelled
of 100 m?
v^{2}  u^{2} = 2as,
v^{2} = 2as + u^{2},
v = √(2as + u^{2})
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?
v^{2}  u^{2} = 2as,
v^{2} = 2as + u^{2},
v = √(2as + u^{2})
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) +
(17.3^{2})} = √(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/s^{2}.
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?
v^{2}  u^{2} = 2as,
u^{2} = 2as  v^{2}, u^{2} = 2as
+ v^{2},
u =
√(v^{2}  2as)
Initially both the launch aircraft and
the rocket plane are travelling at the same speed u.
v = 300 m/s, a = 10 m/s^{2}, s =
1875 m
u = √(v^{2}  2as) =
√{(300^{2})  (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.
v^{2}  u^{2} =
2as, 2as = v^{2}  u^{2},
a = (v^{2}  u^{2})
÷ 2s
s = 400 m, u = 20 m/s, v = 30 m/s (u and
v being the initial and final velocities)
a = (30^{2}  20^{2}) /
(2 x 400) = (900  400) / 800 = 0.625 =
0.63 m/s^{2}
(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/s^{2}.
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) v^{2}  u^{2} = 2as,
2as = v^{2}  u^{2},
s
= (v^{2}  u^{2}) ÷ 2a
a 4.00 m/s^{2}, u = 10.0 m/s, v = 200 m/s
(u and v being the initial and final velocities)
s = (200^{2}  10^{2}) /
(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/s^{2}, at what
theoretical velocity does the object hit the ground?
v^{2}  u^{2} = 2as,
v^{2} = 2as + u^{2},
v = √(2as + u^{2})
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 + u^{2}) =
√(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.
v^{2}  u^{2} = 2as,
rearranging s
= (v^{2}  u^{2}) ÷ 2a
s = height, v = 10 m/s, u = 0 m/s, a
= 9.8 m/s^{2} (gravitational acceleration near Earth's surface)
height = s = (10^{2}
 0^{2}) ÷ (2 x 9.8) = 100 ÷ 19.6 =
5.1 m
Q3.7
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