Question 106: Motion in 1D

Question

A particle moves in a straight line so that its displacement $$x$$ at any time $$t$$ is given by $$x^2=1+t^2$$. Its acceleration at any time $$\mathrm{t}$$ is $$x^{-\mathrm{n}}$$ where $$\mathrm{n}=$$ _________.

Accepted Answer

Given the displacement of the particle $$x^2 = 1 + t^2$$, we want to find the acceleration, which is the second derivative of displacement with respect to time, $$a = \frac{d^2x}{dt^2}$$, and we are given that the acceleration at any time $$t$$ is $$x^{-n}$$, for us to find the value of $$n$$.
First, let's find the first derivative of displacement with respect to time, which gives us the velocity. Differentiating $$x^2 = 1 + t^2$$ with respect to t, we get:
$$2x\frac{dx}{dt} = 2t$$
This simplifies to:
$$\frac{dx}{dt} = \frac{t}{x}$$
Now, let's differentiate this velocity to find the acceleration:
$$a = \frac{d^2x}{dt^2} = \frac{d}{dt}\left(\frac{t}{x}\right)$$
To differentiate $$\frac{t}{x}$$ with respect to $$t$$, we'll use the quotient rule:
$$\frac{d}{dt}\left(\frac{t}{x}\right) = \frac{x\cdot 1 - t\cdot \frac{dx}{dt}}{x^2}$$
Substitute $$\frac{dx}{dt} = \frac{t}{x}$$ into the equation:
$$a = \frac{x(1) - t(\frac{t}{x})}{x^2} = \frac{x - \frac{t^2}{x}}{x^2} = \frac{x^2 - t^2}{x^3}$$
Recalling that the displacement equation given was $$x^2 = 1 + t^2$$, substitute this into our expression for acceleration:
$$a = \frac{1 + t^2 - t^2}{x^3} = \frac{1}{x^3}$$
Thus, the acceleration of the particle at any time $$t$$ is $$x^{-3}$$, which means our value for $$n$$ is $$3$$.

Motion in 1D

A particle moves in a straight line so that its displacement $x at any time t is given by x^2=1+t^2. Its acceleration at any time \mathrm{t} is x^{-\mathrm{n}} where \mathrm{n}=$ _________.

Integer / Numeric Answer Type