The instantaneous velocity of a particle moving in a straight line is given as v = alpha*t + beta*t^

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

That equation is strictly valid ONLY when the acceleration is constant. In this problem, if you differentiate the velocity $v = \alpha t + \beta t^2$, you get acceleration $a = \alpha + 2\beta t$. Since acceleration depends on time $t$, it is not constant, and the standard equations of motion cannot be used. You must use calculus.

Q: Does this integral give distance or displacement?

Technically, the integral of velocity gives displacement. However, since time $t$ is positive (from 1s to 2s) and assuming $\alpha$ and $\beta$ are positive constants, the velocity $v$ is always positive. This means the particle never turns back. When a particle moves in a straight line without reversing direction, the magnitude of displacement is exactly equal to the distance travelled.

Q: What does the area under the velocity-time graph represent?

The area under the velocity-time graph between two time points represents the displacement of the particle during that time interval. This is the geometric interpretation of the definite integral we just solved.

Q: How do I know when to differentiate and when to integrate in kinematics?

Think of a ladder: Displacement -> Velocity -> Acceleration. To go down the ladder (e.g., finding velocity from displacement), you differentiate with respect to time. To go up the ladder (e.g., finding displacement from velocity), you integrate with respect to time.

Q: What happens to the constants $\alpha$ and $\beta$ during integration?

According to the constant multiple rule in calculus, constants that are multiplying a function can be pulled out of the integral. So, $\int \alpha t \, dt$ becomes $\alpha \int t \, dt$. They just 'come along for the ride' while you integrate the variable $t$.

Question

The instantaneous velocity of a particle moving in a straight line is given as $v = \alpha t + \beta t^2$, where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between $1	ext{s}$ and $2	ext{s}$ is

Options:

(A) $3\alpha + 7\beta$

(B) $\frac{3}{2}\alpha + \frac{7}{3}\beta$

(C) $\frac{\alpha}{2} + \frac{\beta}{3}$

(D) $\frac{3}{2}\alpha + \frac{7}{2}\beta$

Sigma Percentile · Accepted Answer

### The Calculus of Motion

Imagine you are tracking a particle moving along a straight line. Its velocity isn't constant; it's changing every single second. The problem gives us the exact mathematical rule governing this velocity: $v = \alpha t + \beta t^2$. This is a quadratic equation, meaning if we were to graph velocity against time, we would see a beautiful parabolic curve sweeping upwards.

Our mission is to find the total distance this particle travels between the $1^{	ext{st}}$ and $2^{	ext{nd}}$ second.

### The Master Equation

How do we extract distance from velocity? We must turn to the fundamental definitions of kinematics. Velocity is defined as the rate of change of displacement with respect to time. Mathematically, this is written as a derivative:

$$v = \frac{ds}{dt}$$

To find the small displacement $ds$ over an infinitesimally small time interval $dt$, we rearrange the equation:

$$ds = v \, dt$$

To find the total distance $s$ over a macroscopic time interval, we must sum up all these tiny $ds$ segments. In the language of calculus, this continuous summation is integration. Geometrically, this integral represents the exact area under the velocity-time graph between our two time limits.

$$s = \int_{t_1}^{t_2} v \, dt$$

### Executing the Integral

Now, we substitute our specific velocity function into the integral and set our limits from $t = 1$ to $t = 2$:

$$s = \int_{1}^{2} (\alpha t + \beta t^2) \, dt$$

Integrating this polynomial is straightforward. We apply the power rule of integration, $\int t^n dt = \frac{t^{n+1}}{n+1}$. The constants $\alpha$ and $\beta$ simply multiply the results.

$$s = \left[ \alpha \frac{t^2}{2} + \beta \frac{t^3}{3} ight]_{1}^{2}$$

### Final Calculation

The final step is to evaluate this expression at the upper limit ($t=2$) and subtract the value at the lower limit ($t=1$). This is where we must be careful with our arithmetic to avoid silly mistakes.

$$s = \left( \alpha \frac{2^2}{2} + \beta \frac{2^3}{3} ight) - \left( \alpha \frac{1^2}{2} + \beta \frac{1^3}{3} ight)$$

Let's group the $\alpha$ and $\beta$ terms together for cleaner algebra:

$$s = \frac{\alpha}{2} [2^2 - 1^2] + \frac{\beta}{3} [2^3 - 1^3]$$

$$s = \frac{\alpha}{2} [4 - 1] + \frac{\beta}{3} [8 - 1]$$

$$s = \frac{3\alpha}{2} + \frac{7\beta}{3}$$

And there we have it! By applying the fundamental theorem of calculus to our kinematic definitions, we have successfully derived the exact distance travelled by the particle. **The correct option is (b).**

Animated Solution for Physics - Kinematics: The instantaneous velocity of a particle moving in a straight line is given as $v = α t + β t^{2}$ , where $α$ and $β$ are constants. The distance travelled by the particle between $1 s$ and $2 s$ is

Select Answer:

Visualized Solution

Visualizing the Motion

Linking Velocity and Distance

Setting up the Integral

Performing the Integration

Applying the Limits

Simplifying the Expression

Final Answer

The Way Forward

The Sigma Insight: Motion in a Straight Line

The Calculus of Motion

The Master Equation

Executing the Integral

Final Calculation

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