# 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

## 1 Expert Answer

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.

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.

Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

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