Further Mathematics Questions and Answers

Given, \(a(t) =  (3t - 2) ms^{-2}\), the first integration of a(t), with respect to t, gives v(t) (the velocity). The second integration of a(t) or first integration of v(t) gives s(t).

\(v(t) = \int (3t - 2) \mathrm {d} t = \frac{3}{2}t^{2} - 2t\)

\(s(t) = \int (\frac{3}{2}t^{2} - 2t) \mathrm {d} t\)

= \(\frac{t^{3}}{2} - t^{2}\)

When t = 3s,

\(s(3) = \frac{3^{3}}{2} - 3^{2} = \frac{27}{2} - 9 \)

= \(\frac{9}{2}\)

Given \(a(t) = 3t - 2\), \(v(t) = \int (a(t)) \mathrm {d} t\)

= \(\int (3t - 2) \mathrm {d} t = \frac{3}{2}t^{2} - 2t\)

\(\frac{3}{2}t^{2} - 2t = 0 \implies t(\frac{3}{2}t - 2) = 0\)

\(t = \text{0 or t} = \frac{3}{2}t - 2 = 0 \implies t = \frac{4}{3}\)

The time t = 0 was the starting point, The next time v = 0 m/s is at \(t = \frac{4}{3} seconds\).

No explanation has been provided for this answer.

\(\frac{x^{3} + 5x + 1}{x^{3}} \equiv  1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}\)

\(\equiv \int (1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}) \mathrm {d} x = \int (1 + 5x^{-2} + x^{-3}) \mathrm {d} x\)

= \((x - 5x^{-1} - \frac{1}{2}x^{-2} + c)\)

= \(x - \frac{5}{x} - \frac{1}{2x^{2}} + c\).

\(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x  \end{vmatrix} =  -3 \implies (1+2x)(3-x) - (-6) = -3\)

\(3 - x + 6x - 2x^{2} + 6 = -3\)

\(-2x^{2} + 5x + 3 + 6 + 3 = 0\)

Multiplying through with -1, 

\(2x^{2} - 5x -12 = 0\)

\((2x + 3)(x - 4) = 0 \implies x = \frac{-3}{2} , 4\)