The vertical velocity vvv of a test projectile, in m s−1\text{m s}^{-1}m s−1, is modelled by the function v(t)=t(21−t)v(t) = \sqrt{t}(21 - t)v(t)=t(21−t) for t≥0t \ge 0t≥0, where ttt is the time in seconds after launch.
Find, using calculus, the time ttt at which the projectile reaches its maximum vertical velocity.
A graph of vvv against ttt shows a region R1R_1R1 bounded by the curve and the ttt-axis between t=0t = 0t=0 and the point where the velocity first returns to zero (t=21t = 21t=21). A second region R2R_2R2 is bounded by the curve, the ttt-axis, and the vertical line t=Kt = Kt=K, where K>21K > 21K>21.
Given that the area of R1R_1R1 is equal to the area of R2R_2R2, use calculus to determine the exact value of KKK.
363 exam-style questions on CCEA A Level Maths 3.6 Integration (A-level only), covering 3.6.1 Integration (A-level only), 3.6.2 Integration (A-level only), 3.6.3 Integration (A-level only), 3.6.4 Integration (A-level only), 3.6.5 Integration (A-level only), 3.6.6 Integration (A-level only), and 3.6.7 Integration (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.