A force whose time dependence is shown in Figure III acts on…

A force whose time dependence is shown in Figure III acts on a body of mass $m = 100\ \mathrm{kg}$. Draw the velocity and acceleration diagrams of the body if (a) it was initially at rest, (b) it initially moved at $v = 10\ \mathrm{m/s}$ opposite to the force.

Explanation of Method

Velocity and acceleration diagrams from a force diagram.

Step 1: The method

At every moment Newton's second law gives the acceleration, and the velocity is the initial velocity plus the area under the acceleration graph: $a(t) = \frac{F(t)}{m}$.

\[ v(t) = v_0 + \int_0^t a\,dt'. \]

The acceleration diagram is the force diagram with the vertical scale divided by $m = 100\ \mathrm{kg}$.
The velocity diagram rises where $a \gt 0$, falls where $a \lt 0$, is straight where $a$ is constant and curved where $a$ changes linearly.
Case (b) is case (a) shifted down by 10 m/s (taking the force direction as positive, $v_0 = -10\ \mathrm{m/s}$).

Step 2: Worked example (assumed force graph)

Figure III is not available, so here is the procedure for the force shape sketched in the original solution, with assumed values: $F = 200\ \mathrm{N}$ for 0–2 s, then falling linearly to 0 at 4 s and to $-100\ \mathrm{N}$ at 6 s.

\[ a = 2\ \mathrm{m/s^2} \ (0\text{–}2\ \mathrm{s}), \]

\[ a: 2 \to 0 \ (2\text{–}4\ \mathrm{s}), \]

\[ a: 0 \to -1\ \mathrm{m/s^2} \ (4\text{–}6\ \mathrm{s}). \]

Adding up the areas (4, then 2, then −1 m/s):

t (s)	0	2	4	6
(a) v (m/s)	0	4	6	5
(b) v (m/s)	−10	−6	−4	−5

$a\ (\mathrm{m/s^2})$ $t\ (\mathrm{s})$ $2$ $4$ $6$ $-1$ $1$ $2$ $v\ (\mathrm{m/s})$ $t\ (\mathrm{s})$ $2$ $4$ $6$ $-10$ $-6$ $-4$ $4$ $6$ $a = F/m$ $\text{(a)}$ $\text{(b)}$ $\text{example: } F = 200\ \mathrm{N} \to 0 \to -100\ \mathrm{N}$

(a) from rest
(b) starting at −10 m/s

Worked for an assumed force graph (Figure III is not available): 200 N for 2 s, falling linearly to 0 at 4 s and to −100 N at 6 s. The acceleration graph has the shape of the force graph; the velocity is the running area under it, shifted by the initial velocity.

Result

The acceleration diagram is $a = F/m$; the velocity is the running area under $a(t)$, shifted by $v_0$. For the assumed force, (a) $v = 0,\,4,\,6,\,5\ \mathrm{m/s}$ and (b) $v = -10,\,-6,\,-4,\,-5\ \mathrm{m/s}$ at $t = 0,\,2,\,4,\,6\ \mathrm{s}$.

Remark

About the original solution. The figure with the actual force is missing, so the original drew an invented example; its velocity diagram (0 → 4 → 5 → 3) does not match its own acceleration graph, which gives 0 → 4 → 6 → 5. It also claimed that in (b) the body reverses direction; in this example it only slows from 10 to 4 m/s and never turns around. Redraw the diagrams with the real values from Figure III using the same steps.