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Question

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Answer 1

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-t²/7) * 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/sec²

Answer 2

a) v(t) = v₀*e^^-1/2 = 12*e^^-t/2

b) dv/dt = - (v_0/ t_o)*e^^-t/t0

1/v dv/dt|_t=2 = - (v_0/ t_o)*e^^{-t/ t}₀ / v₀*e^^-t/t₀ = 1/2

1/v dv/dt|_t=4 = 1/2

c) dv/dt = - 6*e^^-t/t₀

t = 2s  dv/dt = -6e^-1 = - 6/e

    t = 4s  dv/dt = -6e^-2 = - 6/e²

d) Gradient is dv/v

e) d²V/dt² = 3*e^^-t/t₀

Answer 3

I'm going to keep the "out" part out so that

G=20log(10V)

We know that generally,

d/dx[log(x)]=1/xln(a)

but we must apply the chain rule, or the "derivative of the outside times the derivative of the inside" rule.

That is, we have multiply the derivative of log with the derivative of the stuff inside the parentheses, where

dG/dV=(20/V*ln(10))*(10)=30/V*ln(10)=(30/ln(10))*(V^-1)

I purposefully wrote it in an unaesthetic way, but now the second derivative is just a power rule question