Calculating velocity from a Force-Time graph

PROBLEM:

A force F acts on a stationary object of a mass of 96 g for a duration of 30 milliseconds, causing it to move away from its initial position in a straight line on a frictionless surface. Fig. 1 shows the force as a function of time during that 30 milliseconds. Calculate the velocity of the object at times t = 5 ms, 20 ms, and 35 ms.

Fig. 1

SOLUTION:

Let $m$ be the mass of the object, and $\textbf{v}_{0}$ , $\textbf{v}_{5}$ , $\textbf{v}_{20}$ , $\textbf{v}_{35}$ be its velocity at time $t=$ 0 s, 5 s, 20 s, 35 s, respectively.
The force $\textbf{F}$ that acts on the object during time $t$ provides the object with an impulse of $\textbf{F}t$ and change its momentum from $\textbf{p}_{1}$ to $\textbf{p}_{2}$ . According to the impulse-momentum theorem, $\text{Impulse}=\text{change in momentum}$ , we have:

$\texrm{(1)}\hspace{20px}\displaystyle{\textbf{F}t = \textbf{p}_{2} - \textbf{p}_{1} = m\textbf{v}_{2} - m\textbf{v}_{1}}$

As the object moves in a straight line, we can use this line as $x$ -axis with forward direction as positive, and make it an one dimensional motion. As a result, all quantities can be handled as scalars to simplify our calculations. Also, as force $\textbf{F}$ changes with time, it is more appropriate to use differentiation notion for impulse and momentum:

$\texrm{(2)}\hspace{20px}\displaystyle{F(t)dt = mdv}$

where $F(t)dt$ is the impulse on the object during the period of time $dt$ that causes its velocity to change an amount of $dv$ . Velocity $v$ at time $t$ can be calculated by integrating both sides of Eq. (2). Note that $v_{0} =$ 0 as the object is stationary at start; also note that $\int^{t_{2}}_{t_{1}} F(t)dt$ is the area underneath the line $F(t)$ , encapsulated by lines $F =$ 0, $F = F(t)$ , $t = t_{1}$ , and $t = t_{2}$ .

Calculations

${\huge \textbf{$v_5:$}}$

Using SI units, $m=\text{0.096 kg}$ , and $t_0$ , $t_5$ , $t_{20}$ , $t_{35} =$ 0 s, 0.005 s, 0.020 s, 0.035 s, respectively. Magnitude of force $\bf{F}$ at $t_0$ , $t_5$ , $t_{20}$ , and $t_{35}$ are $F_0$ , $F_5$ , $F_{20}$ , and $F_{35}$ , respectively.

$\begin{alignat*}{2} \int^{v_{5}}_0 mdv &= {\int^{0.005}_{0} F(t)dt}\\ {m(v_{5} - v_{0})} &= {\text{the area of the blue triangle in Fig. 2 }}\\ {mv_{5}} &= {\frac{1}{2} \left(\text{0.005} \times F_{5}\right) + v_{0}}\\ {v_{5}} &= {\frac{1}{m} \left(\frac{\text{0.005} \times F_{5}}{2} \right)}\\ {} &= {\frac{1}{0.096} \left(\frac{\text{0.005} \times \text{150}}{2} \right)}\\ {} &= {\text{3.9 m/s}} \end{align*}$

Value of $F_{5}$ is obtained as explained in the “Notes” Section below.

Fig. 2

${\huge \textbf{$v_{20}:$}}$

$\begin{alignat*}{2} \int^{v_{20}}_{v_{5}} mdv &= {\int^{0.020}_{0.005} F(t)dt}\\ {m(v_{20} - v_{5})} &= {\text{orange trapezoid area + yellow trapezoid area (see Fig. 2) }}\\ {m(v_{20} - v_{5})} &= {\frac{1}{2} (F_{5} + F_{10})(\text{0.010} - \text{0.005}) + \frac{1}{2} (F_{10} + F_{20})(\text{0.020} - \text{0.010})}\\ {v_{20} - v_{5}} &= {\frac{1}{m} \left(\frac{1}{2} (\text{150} + \text{300}) \times 0.005 + \frac{1}{2} (\text{300} + \text{150}) \times 0.010 \right)}\\ {v_{20}} &= {\frac{1}{0.096} \left(\text{1.125} + \text{2.25} \right) + v_{5}}\\ {} &= {\text{39.06 m/s}} \end{align*}$

${\huge \textbf{$v_{35}:$}}$

After 0.030 seconds, there is no force acting on the object, therefore it continues to travel with constant speed $v_{30}$ , that is, $v_{35} = v_{30}$ , and $v_{30}$ can be calculated as follows

$\begin{alignat*}{2} \int^{v_{30}}_{v_{20}} mdv &= {\int^{0.030}_{0.020} F(t)dt}\\ {m(v_{30} - v_{20})} &= {\text{area of the green triangle in Fig. 2 }}\\ {m(v_{30}-v_{20})} &= {\frac{1}{2} \left(\text{0.010} \times F_{20}\right)}\\ {v_{30}} &= {\frac{1}{m} \left(\frac{\text{0.010} \times F_{20}}{2}\right)+ v_{20}}\\ {} &= {\frac{1}{0.096} \left(\frac{\text{0.010} \times \text{150}}{2}\right)+v_{20}}\\ {} &= {\text{46.78 m/s}} \end{align*}$

$\textbf{Notes}$
As shown in Fig. 1, $F_{10} =$ 300 N. $F_{5}$ and $F_{20}$ can be calculated using triangle similarity rules. As the blue triangle and the triangle made up by blue triangle and orange trapezoid are similar, we have

$\begin{align*} \frac{F_{5}}{F_{10}} &= {\frac{\text{5}}{\text{10}}}\\ {F_{5}} &= {\left(F_{10} \times \frac{\text{5}}{\text{10}}\right)}\\ {} &= {\text{150 N}} \end{align*}$

Similarly, the green triangle and the triangle made up by green triangle and yellow trapezoid are similar, therefore

$\begin{align*} \frac{F_{20}}{F_{10}} &= {\frac{\text{10}}{\text{20}}}\\ {F_{20}} &= {\left(F_{10} \times \frac{\text{10}}{\text{20}}\right)}\\ {} &= {\text{150 N}} \end{align*}$