Newton's Second Law Along an Axis used forces that stay the same. Real pushes rarely do: a motor ramps up, a gust fades, a bat touches a ball for a few milliseconds. Newton's second law still holds at every instant, so the acceleration simply follows the force, and Velocity and Position from Acceleration does the rest.
Key idea
With a net force $F(t)$ on a mass $m$: $$a(t) = \dfrac{F(t)}{m}, \qquad v(t) = v_0 + \dfrac{1}{m}\int_0^t F(t')\,dt'.$$ The change in velocity is the area under the force–time graph divided by the mass.
Why it works
$F = ma$ is a statement about one instant: the net force now sets the acceleration now. Over a tiny time $dt$ the velocity changes by $dv = a\,dt = \dfrac{F}{m}\,dt$. Adding up all those small changes is an integral. The quantity $\int F\,dt$ is the impulse, which you'll meet again in Impulse as an Integral.
Worked example: A force that grows
A $3\ \text{kg}$ cart starts from rest on a frictionless track, pushed by $F(t) = 6t$ (N, s). Find its speed and position at $t = 4\ \text{s}$.
- $a(t) = F/m = 2t\ \text{m/s}^2$.
- $v(t) = \int_0^t 2t'\,dt' = t^2$, so $v(4) = 16\ \text{m/s}$.
- $x(t) = \int_0^t t'^2\,dt' = \tfrac{1}{3}t^3$, so $x(4) = \tfrac{64}{3} \approx 21.3\ \text{m}$.
Answer: $v = 16\ \text{m/s}$, $x \approx 21.3\ \text{m}$.
Worked example: A push that fades out
A $5\ \text{kg}$ block on ice starts from rest. A push $F(t) = 40 - 10t$ (N, s) acts until it reaches zero. How fast is the block moving then?
- The force reaches zero at $t = 4\ \text{s}$.
- $$\Delta v = \dfrac{1}{5}\int_0^4 (40 - 10t)\,dt = \dfrac{1}{5}\left[40t - 5t^2\right]_0^4 = \dfrac{80}{5} = 16\ \text{m/s}$$
- Geometrically, the area is a triangle: $\tfrac{1}{2}(4)(40) = 80\ \text{N·s}$.
Answer: $16\ \text{m/s}$.
Common mistake
Using the final force as if it had acted the whole time. For the first cart, the force at 4 s is 24 N; $24 \times 4/3 = 32\ \text{m/s}$, twice the true 16 m/s, because the push was weaker early on. Using the starting force goes wrong the other way. Check: the answer must equal the area under the actual $F$–$t$ graph divided by $m$. A rectangle drawn at the final height is far bigger than the triangle under $F = 6t$.
If several forces act, add them first: $F_{\text{net}}(t) = F_{\text{push}}(t) - f_k$, and integrate the net force. The integral of one force alone gives only that force's share of the change in momentum.