free physics video tutorials for all





Linear Motion


Uniform Acceleration



displacement-time graphs

speed-time graphs








To understand this section you must remember the letters representing the variables:


u - initial speed
v - final speed
a - acceleration(+) or deceleration(-)
t - time taken for the change
s - displacement(distance moved)


It is also important to know the S.I. unitsLe   Système International   d'Unités) for these quantities:


u - metres per second (ms-1)
v - metres per second (ms-1)
a - metres per second per second (ms-2)
t - seconds (s)
s - metres (m)

In some textbooks 'speed' is replaced with 'velocity'. Velocity is more appropriate when direction is important.



Displacement-Time graphs



distance time graph



For a displacement-time graph, the gradient at a point is equal to the speed .



back to top



Speed-Time graphs



speed time graph



For a speed-time graph, the area under the curve is the distance travelled.

The gradient at any point on the curve equals the acceleration.


acceleration as a derivative


Note, the acceleration is also the second derivative of a speed-time function.



back to top



Equations of Motion


One of the equations of motion stems from the definition of acceleration:


acceleration = the rate of change of speed


equation definition for acceleration




v equals u plus at


if we define the distance 's' as the average speed times the time(t), then:


distance equals average velocity times time



u plus v equals 2s divided by t 


rearranging (i


v minus u equals at


subtracting these two equations to eliminate v,


derivation of s=ut+half at squared


It is left to the reader to show that :


v squared minus u squared equals 2as


hint: try multiplying the two equations instead of subtracting



equation summary




Example #1


A car starts from rest and accelerates at 10 ms-1 for 3 secs.
What is the maximum speed it attains?


linear horizontal motion problem #1



Example #2


A car travelling at 25 ms-1 starts to decelerate at 5 ms-2.

How long will it take for the car to come to rest?


linear horizontal  motion problem#2



Example #3


A car travelling at 20 ms-1 decelerates at 5 ms-2.
How far will the car travel before stopping?


linear horizontal motion problem #3




Example #4


A car travelling at 30 ms-1 accelerates at 5 ms-2 for 8 secs.
How far did the car travel during the period of acceleration?


linear horizontal motion problem #4



back to top



Vertical motion under gravity


These problems concern a particle projected vertically upwards and falling 'under gravity'.

In these types of problem it is assumed that:


air resistance is minimal

displacement & velocity are positive(+) upwards & negative(-) downwards

acceleration(g) always acts downwards and is therefore negative(-)

acceleration due to gravity(g) is a constant



Example #1


A stone is thrown vertically upwards at 15 ms-1.

(i) what is the maximum height attained?
(ii) how long is the stone in the air before hitting the ground?

(Assume g = 9.8 ms-2. Both answers to 2 d.p.)


gravity problem #1a



gravity problem #1b




Example #2


A boy throws a stone vertically down a well at 12 ms-1.
If he hears the stone hit the water 3 secs. later,

(i) how deep is the well?
(ii)what is the speed of the stone when it hits the water?

(Assume g = 9.8 ms-2. Both answers to 1 d.p.)


gravity proble #2a



gravity problem #2b



back to top




this week's promoted video


 from Physics Trek



creative commons license

All downloads are covered by a Creative Commons License.
These are free to download and to share with others provided credit is shown.
Files cannot be altered in any way.
Under no circumstances is content to be used for commercial gain.





©copyright 2016 - All Rights Reserved