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Question

maths matbs mathe mathe maths maths maths maths maths maths marns matns marhs marjs

Answer 1

First find the constant A from velocity info, with "tmax" = tmaxspeed = 7: v(2.6) = 28 = A (1 - e^-t/tmax) = A (1 - e^-2.6/7) = A (1 - 0.68975), A = 28/0.31025 = 90.249.

Let u = -t/tmax; rewrite v(t) = A - Ae^u. Acceleration is slope of v(t), so a(t) = dv/dt = (dv/du)(du/dt) = (-Ae^u)(-1/tmax) = (A/7)e^-t/tmax , a(t) = 12.893 e^-t/7. Then a(0) = 12.9 m/s², and asymptote at very large t is zero acceleration, a(∞) = 0.

Answer 2

First find constant A from velocity info: v(2.6) = 28 = A (1 - e^-t/tmax) = A (1 - e^-2.6/7) = A (1 - 0.68975), A = 28/0.31025 = 90.249.

Let u = -t/tmax; du = -(1/tmax) dt or dt = -tmax du. Integrate x = -tmax ∫(A - Ae^u ) du to get x = -tmax A [u - e^u] + C or x = tmax A [t/tmax + e^-t/tmax ] + C = A [t + tmax e^-t/tmax ] + C.

Evaluate x from zero to 10.46: x(10.46) = 90.249 [(10.46 + 7e^-10.46/7) - (0 + 7*1)] + C = 5.0309 A + C = 400, C = 400 - 454.0 = -54.0. Now x(t) = 90.249 (t + 7e^-t/7) - 54.0, x(0) = 90.249*7 -54.0 = 577.7 m. As t gets very large, x keeps increasing but the velocity has an asymptote of 90.25 m/s.