Question

a accelerated body covers distance a,b,c in time l,m,n show that a(m-n)+b(n-l)+c(l-m)=0

Answer 1

An accelerated body covers distance $a, b, c$ in time $l, m, n$, respectively.

Show that $a \cdot (m - n) + b \cdot (n - l) + c \cdot (l - m) = 0$.

Solution: Distance $s$, traveled by accelerated body is equal to $s = v_0 \cdot t + \frac{a \cdot t^2}{2}$, where $a$ is the acceleration of the body, $v_0$ is the initial velocity of the body and $t$ is the time of movement. We assume that initial velocity is zero. Then, $a = \frac{a \cdot l^2}{2}$. After time $l$ velocity of body will be equal to $v_l = a \cdot l$, and then $b = v_l \cdot m + \frac{a \cdot m^2}{2} = a \cdot l \cdot m + \frac{a \cdot m^2}{2} = a \cdot m \cdot (l + m/2)$. After time $m$ velocity of body will be equal to

$$\begin{array}{l} v_m = v_l + a \cdot m = a \cdot l + a \cdot m = a \cdot (l + m), \text{ and then } c = v_m \cdot n + \frac{a \cdot n^2}{2} = a \cdot (l + m) \cdot n + \frac{a \cdot n^2}{2} = \\ = a \cdot n \cdot (l + m + n/2). \end{array}$$

We can substitute these formulas into initial equation:

$$\begin{array}{l} \frac{a \cdot l^2}{2} \cdot (m - n) + a \cdot m \cdot (l + m/2) \cdot (n - l) + a \cdot n \cdot (l + m + n/2) \cdot (l - m) = \\ = a \cdot \left[ \frac{m l^2}{2} - \frac{n l^2}{2} + m n l - m l^2 + \frac{n m^2}{2} - \frac{l m^2}{2} + n l^2 - m n l + m n l - n m^2 + \frac{l n^2}{2} - \frac{m n^2}{2} \right] = \\ = a \cdot \left[ \frac{n l^2}{2} + m n l - \frac{m l^2}{2} - \frac{l m^2}{2} - \frac{n m^2}{2} + \frac{l n^2}{2} - \frac{m n^2}{2} \right] = a \cdot \left[ m n l + \frac{n (l^2 - m^2) - m (l^2 + n^2) + l (n^2 - m^2)}{2} \right] \neq 0 \end{array}$$

As you see, the result of such formula where $0 < l < m < n$ can't be equal to zero. We can make a conclusion that initial formula is false, or that condition of the question isn't full.