Page 188 - Physics

Physics for Secondary Schools

Total area under the curve = area of a But the term in brackets is displacement,
trapezium s (i.e., from the second equation of
motion). Therefore,
1
Areaof trapezium  sum of paralell sides  height 2 u  2 2as
v 
2
1 This is the third equation of motion.
Areaof trapezium  sum of paralell sides  height
2
Alternatively, from the defi nition of
1
A  OA BC  OC distance moved,
2  vu 
1 s    t 
A  uv  t  2 
2
From the fi rst equation of motion, t can
From the fi rst equation, v  at be found as:
u
v  u vu

t  t 
a a
1  vu  
A  uv   
2  a  Substitution gives

1  vu 1  vu    vu v u  

 u    v   s    
2  a  2  a   2  a 


uv u 2 v  2 uv v  2 vu vu u 2

  but A = s s 
2a 2a 2a
2as v 2  u 2 2as v 2  u 2
v  2 u  2 2as
Therefore, v  2 u  2 2as

The third equation of the linear motion Example 8.7
can also be derived algebraically by A car starts from rest and
combining the fi rst two equations and accelerates uniformly at the rate
eliminating t. Using the fi rst equation of 2 m/s for 6 s. It then maintains
2
v u at , square both sides of the a constant speed for 30 seconds.
equation to get: After the brakes are applied, it
decelerates uniformly to rest in 5
v  2 u  2 2uat a t 22 s. Calculate:

Factorise 2a on the right-hand side to get: (a) the total distance covered in

1 metres.
v  2 u  2 2( a ut  at 2 ) (b) the maximum speed reached.
2

182
Student’s Book Form One

Physics Form 1 Final.indd 182 16/10/2024 20:58