Question

A particle is moving such that its distance s from a fixed point at any time t is given by

$s = Asin \ t + Bcos \ t$, where A, B $\in$ R.

Show that the particle's acceleration is always numerically equal to its distance from the fixed point.

Answer 1

The particle's acceleration is always numerically equal to its distance from the fixed point.

Answer 2

To show that the particle's acceleration is always numerically equal to its distance from the fixed point, we need to find the acceleration by differentiating the given distance function with respect to time.

Given the distance $s$ of the particle from a fixed point at time $t$ as: $s = A\sin t + B\cos t$

First, find the velocity ($v$) of the particle by differentiating $s$ with respect to $t$: $v = \frac{ds}{dt}$ $v = \frac{d}{dt}(A\sin t + B\cos t)$ $v = A\cos t - B\sin t$

Next, find the acceleration ($a$) of the particle by differentiating $v$ with respect to $t$: $a = \frac{dv}{dt}$ $a = \frac{d}{dt}(A\cos t - B\sin t)$ $a = -A\sin t - B\cos t$

Now, observe the relationship between the acceleration $a$ and the distance $s$: $a = -(A\sin t + B\cos t)$

Since $s = A\sin t + B\cos t$, we can substitute $s$ into the equation for $a$: $a = -s$

The numerical value (magnitude) of acceleration is $|a|$, and the numerical value (magnitude) of distance is $|s|$. From $a = -s$, we take the absolute value on both sides: $|a| = |-s|$ $|a| = |s|$

This shows that the numerical value of the particle's acceleration is always equal to its distance from the fixed point.

The motion described by $s = A\sin t + B\cos t$ with acceleration $a = -s$ is a characteristic of Simple Harmonic Motion (SHM) where the angular frequency $\omega = 1$.

Explanation of the solution:

1. Differentiate the given position function $s(t)$ once to find the velocity $v(t)$.

2. Differentiate the velocity function $v(t)$ once to find the acceleration $a(t)$.

3. Compare the expression for $a(t)$ with the original expression for $s(t)$.

4. Show that $a(t) = -s(t)$, which implies $|a(t)| = |s(t)|$.