A vertical-launch research drone is programmed to reach a target height of 500 metres. The drone’s vertical velocity, w m s−1w\text{ m s}^{-1}w m s−1, is modelled by the equation:
w=25.6−20e−0.5t−0.04e0.25t w = 25.6 - 20e^{-0.5t} - 0.04e^{0.25t} w=25.6−20e−0.5t−0.04e0.25twhere ttt is the time in seconds after the start of the ascent phase.
Find the maximum vertical velocity of the drone, giving your answer to one decimal place. Fully justify that the value you have found is a maximum.
Find an expression for the height of the drone above its starting position, hhh, in terms of ttt.
In a specific test flight, the drone reached the 500-metre target in exactly 28.0 seconds. Comment on the accuracy of the model.
425 exam-style questions on OCR A Level Maths 1.7 Differentiation, covering 1.7.1 Derivative as gradient of the tangent, 1.7.2 Gradient of the tangent at a point, 1.7.3 Sketching the gradient function, 1.7.4 Second derivatives, 1.7.5 Second derivative as rate of change of gradient, 1.7.6 Convex, concave and points of inflection (A-level only), 1.7.7 Differentiation from first principles for powers of x, 1.7.8 Differentiation from first principles for sin x and cos x (A-level only), 1.7.9 Differentiating x^n, 1.7.10 Differentiating e^(kx) and a^(kx) (A-level only), 1.7.11 Differentiating trigonometric functions (A-level only), 1.7.12 Derivative of ln x (A-level only), 1.7.13 Tangents and normals, 1.7.14 Stationary points, 1.7.15 Increasing and decreasing functions, 1.7.16 Points of inflection (A-level only), 1.7.17 Product and quotient rules (A-level only), 1.7.18 Chain rule (A-level only), 1.7.19 Parametric and implicit differentiation (A-level only), 1.7.20 Constructing differential equations (A-level only), and 1.7 Differentiation. Each one has a worked solution and a mark scheme showing where the marks go.