What is the greatest speed he can reach with an acceleration of 5.00 g before blacking out? A jet fighter pilot wishes to accelerate from rest at $5.00$ G to reach Mach-3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than $5.00\text{s}$ . Use $331\text{m/s}$ for the speed of sound. What is the greatest speed he can reach with an acceleration of $5.00$ G before blacking out? What I have done is: Using the formula of acceleration, $a=\frac{Δv}{Δt}=\frac{Δv}{5s}$ Then, I plugged in 5 G into a. $5.00G=\frac{Δv}{5s}$ I assume that the final velocity is $3(331\text{m/s})$ because the fighter pilot wants to reach Mach-3. So, I have $5.00G=\frac{3(331)−v_0}{5}$. And this is where I am stuck. Can one tell me how to continue from here, or if I am completely wrong?

Ok, I am just going to write this out for completeness's sake.

You are given that

$$v_0 = 0$$

since you are told that the pilot begins from rest. You are then told that he experiences constant acceleration of 5.00 g (since we are dealing with m/s and not ft/s, g = 9.81). Now, using the exact equation from the problem, we can write

$$a = \Delta v/\Delta t = ((v_f - v_0)/\Delta t)$$

Plugging in results in:

$$5.00(9.81) = ((v_f - 0)/(5.00))$$

Which leaves you with a final velocity of $$v_f = 245.25 m/s$$ or, if you are particular about significant digits, $$v_f=245 m/s$$

Notice that this is actually less than Mach one, ultimately suggesting that human beings weren't meant to endure accelerations of 5 g's.