Mechanics (M1): Particle moving on a straight line with constant acceleration (Relationships of the 5 Key Formulae)
For M1, it is essential for you to be able to solve questions that involve particles moving in a straight line with constant acceleration.
The representation of symbols is listed as follow:
S - Displacement (Distance; m)
U - Initial Velocity (Starting Speed; m/s)
V - Final Velocity (Final Speed; m/s)
A - Acceleration (The rate of increase in speed; m/s2)
T - Time (Seconds; s)
There are FIVE key formulae:
1) V = U+A.T
2) A = (V-U)/T
3) S = 1/2.(U+V).T
4) S = U.T+1/2.A.T2
5) V^2 = U2+2.A.S
These equations may vary in different textbooks, but the relationships remain consistent. Now I am going to display how are they related to one another.
EQ1 (Equation 1) is obvious, the final speed V m/s is the sum of the initial speed U m/s and the rate of increase in speed A m/s2 over T seconds.
EQ2 is obtained through rearrangement of EQ1 by subtracting U from V and then divide (V-U) by T.
EQ3 may seem a bit tricky. Remeber U and V are the speed of travel and T is the time the particle has travelled. It is important to notice that U and V are the speed travelled at a different time!
For example, A particle travels at the speed U m/s for (T_1) seconds and then it travels at the speed V m/s form (T_1) second to (T) second. The total time travel is T. If we use (U+V)T = UT + VT instead of U(T_1)+V(T-T_1), then we would have an incorrect result as the speed of U m/s, and V m/s are counted twice. Once it is taken into account, then you would realise (U+V).T/2 gives you the total distance it has travelled.
EQ4 can be derived from EQ3 using EQ1.
EQ3: S = 1/2.(U+V).T
S = 1/2.(U+(U+A.T)).T (Now we substitute V with U+A.T using EQ1: V = U+A.T)
S = 1/2.(2.U+A.T)T
S = U.T+A.T2 (EQ4)
EQ5 is the trickiest one. It will require a bit of time to understand, but it is not difficult. I do recommend you to use a page and a pen to work it out as you follow the explanation.
EQ5: V^2 = U2+2.A.S (Labelled it as EQA)
We know that (using EQ1)
V2 = V.V = (U+A.T).(U+A.T) = U2+2.U.A.T+A2.T2 (Labelled it as EQB)
Now when we compare the two equations (EQA and EQB), we found that
2.A.S = 2.U.A.T+A2.T2 (Labelled it as EQC)
Now if we can show EQC is true then we have proved EQ5 is true!
Since S is presented in EQC, we know we have to include one of the relations where S exists, i.e. EQ3 or EQ4.
Now, 2.U.A.T+A2.T2
= A(2.U.T+A.T2) (Factoring out A; now it looks like EQ4! Let's rearrange it to include EQ4!)
= 2.A.(U.T+1/2.A.T2) (Factoring out 2; now it is 2.A.(EQ4)!)
= 2.A.S
Therefore EQC is true, which means EQ5 is true!
This relationship may not be asked during the exam, but it is beneficial to know. As for exam preparation, I would recommend you to memorise EQ1, EQ3 and EQ5. EQ1 and EQ3 enable you to quickly obtain EQ2 and EQ4. EQ5 may take some valuable times to derive so it would be the best to know it before the exam.
