(i)[2]
Show that $\frac{dv}{dx} = 0.45vx - 1.5$.
(ii)[2]
Find the value of $x$ for which the acceleration of $P$ is zero.
(iii)[5]
Given that the minimum value of $v$ is positive, determine the possible values of the projection speed.
Mathematics 9709 · AS & A Level · Representation of data
Show that $\frac{dv}{dx} = 0.45vx - 1.5$.
Find the value of $x$ for which the acceleration of $P$ is zero.
Given that the minimum value of $v$ is positive, determine the possible values of the projection speed.
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Use Newton’s second law in the horizontal direction: $0.2v\dfrac{dv}{dx} = 0.09\sqrt{x} - 0.3$” …