Solve v=5000(1-t/40)^2

$$v = 5000 ( 1 - \frac { t } { 40 } ) ^ { 2 }$$

Solve for v

$$v=\frac{25\left(40-t\right)^{2}}{8}$$

Short Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1-\frac{t}{40}\right)^{2}.

$$v=5000\left(1+2\left(-\frac{t}{40}\right)+\left(-\frac{t}{40}\right)^{2}\right)$$

Cancel out 40, the greatest common factor in 2 and 40.

$$v=5000\left(1+\frac{t}{-20}+\left(-\frac{t}{40}\right)^{2}\right)$$

Calculate -\frac{t}{40} to the power of 2 and get \left(\frac{t}{40}\right)^{2}.

$$v=5000\left(1+\frac{t}{-20}+\left(\frac{t}{40}\right)^{2}\right)$$

Use the distributive property to multiply 5000 by 1+\frac{t}{-20}+\left(\frac{t}{40}\right)^{2}.

$$v=5000+5000\times \frac{t}{-20}+5000\times \left(\frac{t}{40}\right)^{2}$$

Cancel out -20, the greatest common factor in 5000 and -20.

$$v=5000+250t+5000\times \left(\frac{t}{40}\right)^{2}$$

To raise \frac{t}{40} to a power, raise both numerator and denominator to the power and then divide.

$$v=5000+250t+5000\times \frac{t^{2}}{40^{2}}$$

Express 5000\times \frac{t^{2}}{40^{2}} as a single fraction.

$$v=5000+250t+\frac{5000t^{2}}{40^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 5000+250t times \frac{40^{2}}{40^{2}}.

$$v=\frac{\left(5000+250t\right)\times 40^{2}}{40^{2}}+\frac{5000t^{2}}{40^{2}}$$

Since \frac{\left(5000+250t\right)\times 40^{2}}{40^{2}} and \frac{5000t^{2}}{40^{2}} have the same denominator, add them by adding their numerators.

$$v=\frac{\left(5000+250t\right)\times 40^{2}+5000t^{2}}{40^{2}}$$

Do the multiplications in \left(5000+250t\right)\times 40^{2}+5000t^{2}.

$$v=\frac{8000000+400000t+5000t^{2}}{40^{2}}$$

Calculate 40 to the power of 2 and get 1600.

$$v=\frac{8000000+400000t+5000t^{2}}{1600}$$

Divide each term of 8000000+400000t+5000t^{2} by 1600 to get 5000+250t+\frac{25}{8}t^{2}.

$$v=5000+250t+\frac{25}{8}t^{2}$$

Solve for t

$$t=\frac{2\sqrt{2v}}{5}+40$$

$$t=-\frac{2\sqrt{2v}}{5}+40\text{, }v\geq 0$$