The magnitude of the force (in Newtons) acting on a body varies with time t (in microseconds) as sho

Q: Why do we calculate the area under the graph to find impulse?

Impulse is mathematically defined as the integral of force over time ($J = \int F \, dt$). In calculus, the definite integral of a function represents the area under its graph. Therefore, the area under a Force-Time graph directly gives the total impulse.

Q: Why can't we just use the formula $J = F \times \Delta t$?

The formula $J = F \times \Delta t$ only works when the force is **constant**. In this problem, the force is changing continuously over time (as shown by the sloped lines). When force varies, we must use integration, which geometrically means finding the area under the curve.

Q: Why did we split the area into a trapezium and a triangle?

We split the area to make the calculation easier. The region from $t=4$ to $t=16$ is a complex polygon. By dropping a perpendicular at $t=6$, we divide it into a trapezium (which has a standard area formula) and a triangle (which also has a standard area formula). You could also split the trapezium into a rectangle and a smaller triangle if you prefer!

Q: I got $5000$ as my answer. What did I do wrong?

You likely forgot to convert the units on the time axis! The time is given in microseconds ($\mu\text{s}$), not seconds. You must multiply your time values by $10^{-6}$ to convert them to standard SI units (seconds) before calculating the area. Otherwise, your answer will be $1,000,000$ times too large.

Q: What does the impulse physically represent in this scenario?

Impulse represents the total change in momentum of the body. According to the Impulse-Momentum Theorem ($J = \Delta p$), the $5 \times 10^{-3} \, \text{N-s}$ of impulse we calculated means the body's momentum changed by exactly $5 \times 10^{-3} \, \text{kg}\cdot\text{m/s}$ during that time interval.

Question

The magnitude of the force (in Newtons) acting on a body varies with time $t$ (in microseconds) as shown in the figure. $AB$, $BC$ and $CD$ are straight line segments. The magnitude of the total impulse of the force on the body from $t = 4 \mu	ext{s}$ to $t = 16\mu	ext{s}$ is ......... N-s.

Sigma Percentile · Accepted Answer

Have you ever wondered how we measure the total effect of a force that changes over time? Imagine a tennis racket hitting a ball. The force isn't constant; it starts at zero, peaks rapidly, and drops back to zero. To find the total 'push' or **impulse** delivered to the ball, we can't just multiply force by time. We need to look at the area under the Force-Time graph.

### The Master Equation

In physics, impulse $J$ is defined as the integral of force with respect to time:
$$J = \int F \, dt$$
Geometrically, this integral is exactly equal to the **area under the $F-t$ curve**. In our problem, we are given a specific graph and asked to find the impulse between $t = 4 \, \mu	ext{s}$ and $t = 16 \, \mu	ext{s}$. This means we need to calculate the area bounded by the graph, the x-axis, and the vertical lines at $t = 4$ and $t = 16$.

### Analyzing the Setup

Looking at the graph, the region from $t = 4 \, \mu	ext{s}$ to $t = 16 \, \mu	ext{s}$ isn't a single basic shape. However, we can easily split it into two familiar geometric figures:
1. A **trapezium** ($EBCF$) from $t = 4 \, \mu	ext{s}$ to $t = 6 \, \mu	ext{s}$.
2. A **triangle** ($FCD$) from $t = 6 \, \mu	ext{s}$ to $t = 16 \, \mu	ext{s}$.

By calculating the area of these two shapes separately and adding them together, we will find the total impulse.

### The Microsecond Trap

Before we crunch the numbers, there is a classic trap waiting for us. The time axis is given in **microseconds** ($\mu	ext{s}$), not seconds. If we calculate the area using the raw numbers from the axis, our answer will be off by a factor of a million!

We must convert microseconds to seconds:
$$1 \, \mu	ext{s} = 10^{-6} \, 	ext{s}$$

### Calculating the Areas

Let's start with the trapezium $EBCF$. The formula for the area of a trapezium is $\frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$.
The parallel sides are the force values at $t = 4$ and $t = 6$, which are $200 \, 	ext{N}$ and $800 \, 	ext{N}$. The height is the time interval, $\Delta t = 2 	imes 10^{-6} \, 	ext{s}$.
$$J_1 = \frac{1}{2} 	imes (200 + 800) 	imes (2 	imes 10^{-6})$$
$$J_1 = 1000 	imes 10^{-6} = 10^{-3} \, 	ext{N-s}$$

Next, we calculate the area of the triangle $FCD$. The formula is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.
The base is the time interval from $t = 6$ to $t = 16$, which is $10 	imes 10^{-6} \, 	ext{s}$. The height is the peak force, $800 \, 	ext{N}$.
$$J_2 = \frac{1}{2} 	imes (10 	imes 10^{-6}) 	imes 800$$
$$J_2 = 5 	imes 10^{-6} 	imes 800 = 4000 	imes 10^{-6} = 4 	imes 10^{-3} \, 	ext{N-s}$$

### Final Calculation

The total impulse is simply the sum of the two areas we just calculated:
$$J_{	ext{total}} = J_1 + J_2$$
$$J_{	ext{total}} = 10^{-3} + 4 	imes 10^{-3}$$
$$J_{	ext{total}} = 5 	imes 10^{-3} \, 	ext{N-s}$$

And there we have it! By breaking down a complex varying force into simple geometric areas and carefully managing our units, we've successfully found the total impulse.

Visualized Solution

Impulse and Area under $F - t$ Graph

Identifying the Required Area

Area of Trapezium $E B C F$

Calculating Impulse for $t = 4$ to $6$ $μ s$

Area of Triangle $F C D$

Calculating Impulse for $t = 6$ to $16$ $μ s$

Total Impulse

The Sigma Insight: Inertia, Momentum, and Impulse

The Master Equation

Analyzing the Setup

The Microsecond Trap

Calculating the Areas

Final Calculation

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