The velocity of a particle is v = v_0 + gt + Ft^2. Its position is x = 0 at t = 0, then its displace

Q: Why can't I just use the equation $s = ut + \frac{1}{2}at^2$?

That equation is strictly valid only when acceleration is constant. In this problem, the velocity has a $t^2$ term, which means acceleration ($a = \frac{dv}{dt} = g + 2Ft$) depends on time. Therefore, you must use calculus.

Q: What happens to the constant of integration 'C'?

Because we are using definite integrals with specific limits (from $t=0$ to $t=1$), the constant of integration cancels out. Alternatively, if you use indefinite integration, the initial condition ($x=0$ at $t=0$) will make $C=0$.

Q: Can I just substitute $t=1$ directly into the velocity equation?

No. Substituting $t=1$ into the velocity equation will give you the *velocity* of the particle at $t=1$, not its *displacement*. To find displacement, you must integrate velocity over time.

Q: What does the $F$ term represent physically?

Since $F$ is multiplied by $t^2$ in the velocity equation, it relates to the rate of change of acceleration. In physics, the rate of change of acceleration is called 'jerk'.

Q: Why did we ignore the lower limit during substitution?

The lower limit for time is $t=0$. When you substitute $0$ into the integrated expression $(v_0 t + \frac{gt^2}{2} + \frac{Ft^3}{3})$, every term becomes zero. So, subtracting the lower limit evaluation just means subtracting zero.

Question

The velocity of a particle is $v = v_0 + gt + Ft^2$. Its position is $x = 0$ at $t = 0$, then its displacement after time ($t = 1$) is

Options:

(A) $v_0 + g + F$

(B) $v_0 + \frac{g}{2} + \frac{F}{3}$

(C) $v_0 + \frac{g}{2} + F$

(D) $v_0 + 2g + 3F$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a particle moving along a straight line. We are given its velocity as a function of time: $$v = v_0 + gt + Ft^2$$.

The first thing to notice here is that the velocity is not constant, nor is it changing linearly. Because there is a $$t^2$$ term, the acceleration itself is changing with time. This is a critical observation because it means we **cannot** use the standard equations of motion like $$s = ut + \frac{1}{2}at^2$$. Those equations are strictly reserved for constant acceleration scenarios.

We are also given an initial condition: at $$t = 0$$, the particle is at the origin, so $$x = 0$$. Our goal is to find the position $$x$$ at $$t = 1$$ second.

### The Master Equation
When standard equations fail, we must return to the fundamental definitions of kinematics. How do we connect velocity to position? We know that velocity is the instantaneous rate of change of displacement. Mathematically, this is written as:
$$v = \frac{dx}{dt}$$

To find the displacement $$x$$, we need to isolate $$dx$$ and integrate. Rearranging the equation gives us:
$$dx = v \, dt$$

Now, we set up a definite integral. We integrate the left side from the initial position $$0$$ to the final position $$x$$, and the right side from the initial time $$0$$ to the final time $$1$$:
$$\int_{0}^{x} dx = \int_{0}^{1} (v_0 + gt + Ft^2) \, dt$$

### Final Calculation
Now, we simply execute the integration term by term using the power rule $$\int t^n dt = \frac{t^{n+1}}{n+1}$$.

The integral of the constant $$v_0$$ is $$v_0 t$$. The integral of $$gt$$ is $$\frac{gt^2}{2}$$. And the integral of $$Ft^2$$ is $$\frac{Ft^3}{3}$$. This gives us:
$$x = \left[ v_0 t + \frac{gt^2}{2} + \frac{Ft^3}{3} \right]_{0}^{1}$$

Finally, we substitute the upper limit $$t = 1$$. The lower limit is $$0$$, which will just make the whole expression zero, so we can safely ignore it. Plugging in $$1$$, we get:
$$x = v_0(1) + \frac{g(1)^2}{2} + \frac{F(1)^3}{3}$$

$$x = v_0 + \frac{g}{2} + \frac{F}{3}$$

This elegant expression is our final displacement. Always remember the hierarchy: differentiate to go from position to velocity to acceleration, and integrate to go in the reverse direction!

Animated Solution for Physics - Kinematics: The velocity of a particle is $v = v_{0} + g t + F t^{2}$ . Its position is $x = 0$ at $t = 0$ , then its displacement after time ( $t = 1$ ) is

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Visualized Solution

$v (t) = v_{0} + g t + F t^{2}$

$v = \frac{d x}{d t}$

$\int d x = \int v d t$

$\int t^{n} d t = \frac{t ^{n + 1}}{n + 1}$

$x = v_{0} + \frac{g}{2} + \frac{F}{3}$

$a = \frac{d v}{d t}$

The Sigma Insight: Motion in a Straight Line

Analyzing the Setup Imagine a particle moving along a straight line

The Master Equation When standard equations fail, we must return to the fundamental definitions of kinematics

Final Calculation

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Animated Solution for Physics - Kinematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The velocity of a particle is v=v0​+gt+Ft2. Its position is x=0 at t=0, then its displacement after time (t=1) is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v(t)=v0​+gt+Ft2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=dtdx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫dx=∫vdt

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫tndt=n+1tn+1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=v0​+2g​+3F​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=dtdv​