Question 19: Motion in 1D

Question

The relation between time '$$t$$' and distance '$$x$$' is $$t=\alpha x^2+\beta x$$, where $$\alpha$$ and $$\beta$$ are constants. The relation between acceleration $$(a)$$ and velocity $$(v)$$ is :

Accepted Answer

The relationship between time ($$t$$) and distance ($$x$$) is given by $$t = \alpha x^2 + \beta x$$, where $$\alpha$$ and $$\beta$$ are constants. To find the relation between acceleration ($$a$$) and velocity ($$v$$), we can follow these steps:
First, we differentiate the given equation with respect to time:
$$\begin{aligned} & t = \alpha x^2 + \beta x \quad 	ext{(differentiating with respect to time)} \ & \frac{dt}{dx} = 2\alpha x + \beta \ & \frac{1}{v} = 2\alpha x + \beta \ \end{aligned}$$
Next, we differentiate again with respect to time to find the acceleration:
$$\begin{aligned} & -\frac{1}{v^2} \frac{dv}{dt} = 2\alpha \frac{dx}{dt} \ & 	ext{Since} \quad \frac{dx}{dt} = v, \quad 	ext{we have:} \ & -\frac{1}{v^2} \frac{dv}{dt} = 2\alpha v \ & \frac{dv}{dt} = -2\alpha v^3 \ & 	ext{Therefore,} \quad a = -2\alpha v^3 \end{aligned}$$

Answer

$$a=-5 \alpha v^5$$

Answer

$$a=-3 \alpha v^2$$

Answer

$$a=-2 \alpha v^3$$

Answer

$$a=-4 \alpha v^4$$

Motion in 1D

The relation between time '$t' and distance 'x' is t=\alpha x^2+\beta x, where \alpha and \beta are constants. The relation between acceleration (a) and velocity (v)$ is :

Choose the correct answer: