Submitted by cyuvraj98 on Mon, 02/04/2019 - 14:22
Forums:
2012 STEP 1 Q11
For 2(ii), I cannot wrap my head around why a(P)=0.5(a(A)+a(B))
Where a stands for acceleration and the above expression states acceleration of object P is equal to half the sum of acceleration of objects A and B. The solution for this qn states "If A goes up the slope by 5cm and B goes up by 3cm then P will go down by 0.5(5+3) = 4cm".
Why is this so? How is the displacement then directly proportionally related to acceleration. Is it by the SUVAT eqn s=ut+0.5at^2
Really appreciate your help in this
2012 STEP 1 Q11
$ a_P=0.5(a_A+a_B)$ because the acceleration is linked to how much string passes over the pulley.
If $\delta_A$ length of string passes over pulley $Q$ and $\delta_B$ length of string passes over pulley $R$ then pulley $P$ will move down by a distance of $\frac 1 2 (\delta_A+\delta_B)$. This is what I was trying to say with my example "If A goes up the slope by 5cm and B goes up by 3cm then P will go down by 0.5(5+3) = 4cm".
The reason is linked to the fact that $a=\frac{\text{d}^2 x}{\text {d} t^2}$. For the displacements we have:
\[ x_P = \frac 1 2(x_A+x_B) \]
and then differentiating twice (noting that the constant of $\frac 1 2$ can come outside the derivative and that the derivative of a sum is equal to the sum of the two derivatives; i.e. $\frac{\text{d} kx}{\text {d} t}=k\frac{\text{d} x}{\text {d} t}$ and $\frac{\text{d} (f+g)}{\text {d} t}=\frac{\text{d} f}{\text {d} t}+\frac{\text{d} g}{\text {d} t}$) we have:
\[a_P = \frac 1 2(a_A+a_B) \]