Animated Solution for Physics - Kinematics: The velocity of a particle is $v=v_0+gt+f{t}^2$. If its position is $x=0$ at $t=0$, then its displacement after unit time ($t=1$) is

Imagine you are tracking a particle moving along a straight line. You don't know its exact position at every moment right away, but you have a radar gun that gives you its velocity at any given time $t$. The velocity is given by the equation $v=v_0+gt+f{t}^2$.

This isn't just a simple constant velocity. It's a velocity that changes with time. The $v_0$ term tells us the initial velocity when $t=0$. The $gt$ term tells us that the velocity is increasing linearly, much like an object falling under gravity. The $f{t}^2$ term adds a quadratic curve to the velocity, meaning the acceleration itself is changing!

Our mission is to find the total displacement of this particle after exactly one unit of time, meaning at $t=1$. We are also given a helpful starting condition: at $t=0$, the particle is at the origin, so $x=0$.

How do we bridge the gap between velocity and displacement? We need to look at the fundamental definition of velocity. Velocity is the rate of change of position with respect to time. In the language of calculus, this is written as:

To find the displacement $x$, we need to rearrange this equation to isolate $dx$:

This tiny equation is incredibly powerful. It says that a microscopic change in position ($dx$) is equal to the velocity at that instant multiplied by a microscopic time interval ($dt$). To find the total macroscopic displacement, we need to sum up all these microscopic changes. In calculus, "summing up" continuous changes means taking an integral.

Let's set up our definite integral. We will integrate both sides of our equation. On the left side, we integrate $dx$ from the initial position $x=0$ to the final position $x$. On the right side, we integrate our velocity function with respect to time, from the initial time $t=0$ to the final time $t=1$.

Geometrically, what we are doing here is finding the area under the velocity-time curve between $t=0$ and $t=1$. The area under a $v-t$ graph always represents the displacement.

Now, let's roll up our sleeves and do the math. We will integrate the polynomial term by term using the power rule for integration, $\int t^{n}dt=\frac{t^{n+1}}{n+1}$.

The integral of the constant $v_0$ is $v_{0}t$.

The integral of $gt$ is $g\frac{t^2}{2}$.

The integral of $f{t}^2$ is $f\frac{t^3}{3}$.

Putting it all together and applying our limits from $0$ to $1$:

The final step is to substitute our upper and lower limits into the integrated expression. We plug in $t=1$ first, and then subtract the expression evaluated at $t=0$.

Since every term in the lower limit evaluation contains a zero, the entire second bracket simply vanishes. We are left with:

And there we have it! This is the exact displacement of the particle after one unit of time. By understanding the relationship between velocity and position through calculus, we transformed a dynamic, changing velocity into a clean, precise distance.