Matthew S. answered • 02/26/20

PhD in Mathematics with extensive experience teaching Calculus

We're told that v(2.6) = 28. On the other hand v(2.6) = A(1 - e^{-2.6/7}). Here I'm using the fact that t_{maxspeed}=7. Therefore A = 28 / (1 - e^{-2.6/7}) ≈ 90.25

Since e^{-t/tmaxspeed} is decreasing, v is increasing. So, from the way the problem is phrased, max velocity is at t = 7, and v(7) ≈ 90.25 * (1 - e^{-1}) = 57.05 m/s.

Acceleration a is dv/dt ≈ 90.25 * (-t/7) * (-e^{-t/7}) = 90.25te^{-t/7}/7

To maximize acceleration (call it a), we take its derivative and set to 0: da/dt = (90.25/7) * (1-t^{2}/7) * e^{-t/7}. Critical point is at t = √7. Note that A is increasing on the interval (0, √7) and decreasing for t > √7.

To obtain maximum value, evaluate A at the critical value t = √7:

If I've done the calculation right, a(√7) ≈ (90.25 * √7/7) * (e^{-√7/7}) ≈ 23.7 m/sec^{2}