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

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

## 2 Answers By Expert Tutors

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/s^{2}, and asymptote at very large t is zero acceleration, a(∞) = 0.

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.

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Stanton D.

Hi Matt M., Doesn't look so bad. Consider what that velocity function must look like: starts at 0 for t=0, and you can solve directly to get A out from the 2.6 sec = 28 m/s boundary condition. Then differentiate the v(t)/dt to get a(t) equation, and plug in to find a(0). Note that the "tmaxspeed" designation is a little deceptive, since the asymptote for velocity is approached but never reached! "t(1/e)approachtomaxspeed" might be 'truth in advertising'.02/28/20