Question 1

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

Solution

Solution & Explanation

1. Express Velocity as a Differential Equation

Velocity ($$v$$) is defined as the instantaneous rate of change of displacement ($$x$$) with respect to time ($$t$$):

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

We substitute the given velocity expression into this definition to form our differential equation:

$$\frac{dx}{dt} = v_0 + g \cdot t + f \cdot t^2$$

Separating the variables to prepare for integration gives:

$$dx = (v_0 + g \cdot t + f \cdot t^2) \cdot dt$$

2. Perform Definite Integration with Limits

We now integrate both sides using the boundary conditions specified in the problem statement:

At the initial boundary condition, the time is $$t = 0$$ and the position is $$x = 0$$.
At the final boundary condition, the time reaches unit time $$t = 1$$ and the position reaches $$x = x$$.

Setting up the integration limits:

$$\int_{0}^{x} dx = \int_{0}^{1} (v_0 + g \cdot t + f \cdot t^2) \cdot dt$$

Applying the standard power rule for calculus integration ($$\int t^n \cdot dt = \frac{t^{n+1}}{n+1}$$):

$$[x]_{0}^{x} = \left[ v_0 \cdot t + \frac{g \cdot t^2}{2} + \frac{f \cdot t^3}{3} \right]_{0}^{1}$$

3. Substitute Limits to Find the Final Displacement

Evaluating the limits by substituting the upper value ($$1$$) and subtracting the lower value ($$0$$):

$$x - 0 = \left( v_0 \cdot (1) + \frac{g \cdot (1)^2}{2} + \frac{f \cdot (1)^3}{3} \right) - (0)$$

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

Concept Check: Because acceleration is not constant (due to the time-dependent terms $$g \cdot t$$ and $$f \cdot t^2$$), standard algebraic equations of motion cannot be used. Continuous calculus integration over the entire time interval correctly accounts for the changing acceleration components.