A force F = kt^2 with k = 10 N/s^2 acts on a body of mass m…

A force $F = kt^2$ with $k = 10\ \mathrm{N/s^2}$ acts on a body of mass $m = 1\ \mathrm{kg}$, initially at rest, for $\tau = 3\ \mathrm{s}$. (a) What is the body's momentum when the force stops? (b) How far does it travel while the force acts?

Explanation of Method

A force growing as t².

Step 1: (a) Momentum = impulse

\[ p = J = \int_0^\tau kt^2\,dt = \frac{k\tau^3}{3} = \frac{10 \cdot 27}{3}. \]

\[p = 90\ \mathrm{kg\,m/s} \ \ (v = 90\ \mathrm{m/s})\]

Step 2: (b) Distance

\[ v(t) = \frac{kt^3}{3m}, \]

\[ s = \int_0^\tau v\,dt = \frac{k\tau^4}{12m} = \frac{10 \cdot 81}{12}. \]

\[s = 67.5\ \mathrm{m}\]

$F\ (\mathrm{N})$ $t\ (\mathrm{s})$ $0$ $1$ $2$ $3$ $30$ $60$ $90$ $J = \int_0^3 kt^2\,dt = 90\ \mathrm{N\,s}$ $F = kt^2$

Force against time; the shaded area is the impulse, 90 N·s, which becomes the body's momentum (so $v = 90\ \mathrm{m/s}$ at the end).

Result

\[\boxed{\,p = 90\ \mathrm{kg\,m/s} \ \ (v = 90\ \mathrm{m/s})\,}\]

\[\boxed{\,s = 67.5\ \mathrm{m}\,}\]

Remark

Check. The original results are correct. ("Impulse of the body" means its momentum, which equals the impulse of the force.)