Question

A point moves with constant acceleration and ${v_1}$, ${v_2}$ and ${v_3}$ denoted the average velocities in the three successive intervals ${t_1}$, ${t_2}$, and ${t_3}$ of time. Which of the following relations is correct?
(A) $\dfrac{{{v_1} - {v_2}}}{{{v_2} - {v_3}}} = \dfrac{{{t_1} - {t_2}}}{{{t_2} + {t_3}}}$
(B) $\dfrac{{{v_1} - {v_2}}}{{{v_2} - {v_3}}} = \dfrac{{{t_1} - {t_2}}}{{{t_1} - {t_3}}}$
(C) $\dfrac{{{v_1} - {v_2}}}{{{v_2} - {v_3}}} = \dfrac{{{t_1} - {t_2}}}{{{t_2} - {t_3}}}$
(D) $\dfrac{{{v_1} - {v_2}}}{{{v_2} - {v_3}}} = \dfrac{{{t_1} + {t_2}}}{{{t_2} + {t_3}}}$

Answer 1

Since the acceleration is constant, the average acceleration from first interval to second interval, and from second interval to third interval are equal. Average acceleration can also be written as initial minus final velocity as long as the time is also initial minus final time.

Formula used: In this solution we will be using the following formula;
$a = \dfrac{{{v_f} - {v_i}}}{{{t_f} - {t_i}}}$ where $a$ is the acceleration of a body, ${v_f}$ is the final velocity, ${v_i}$ is the initial velocity, ${t_f}$ is time at final velocity, and ${t_i}$ is the time as initial velocity.

Complete step by step answer
A point is said to move with constant acceleration, during such acceleration, it possess velocity ${v_1}$ at time ${t_1}$, velocity ${v_2}$ at time ${t_2}$, and ${v_3}$ at time ${t_3}$.
Now acceleration between ${v_1}$ and ${v_2}$ will be given as
$a = \dfrac{{{v_f} - {v_i}}}{{{t_f} - {t_i}}} = \dfrac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$ ( making ${v_2}$ as the final velocity) where ${v_f}$ is the final velocity, ${v_i}$ is the initial velocity, ${t_f}$ is time at final velocity, and ${t_i}$ is the time as initial velocity
Now, we can multiply both numerator and denominator as -1, hence
${a_{12}} = \dfrac{{ - \left( {{v_2} - {v_1}} \right)}}{{ - \left( {{t_2} - {t_1}} \right)}} = \dfrac{{{v_1} - {v_2}}}{{{t_1} - {t_2}}}$
Similarly for acceleration between ${v_2}$ and ${v_3}$, it can be written as
${a_{23}} = \dfrac{{ - \left( {{v_3} - {v_2}} \right)}}{{ - \left( {{t_3} - {t_2}} \right)}} = \dfrac{{{v_2} - {v_3}}}{{{t_2} - {t_3}}}$
Now, since the acceleration is said to be constant, the first acceleration and the second acceleration are hence equal, thus
${a_{12}} = {a_{23}}$
$\Rightarrow \dfrac{{{v_1} - {v_2}}}{{{t_1} - {t_2}}} = \dfrac{{{v_2} - {v_3}}}{{{t_2} - {t_3}}}$
By multiplying both sides by ${t_1} - {t_2}$ and dividing both sides by ${v_2} - {v_3}$, we have that
$\dfrac{{{v_1} - {v_2}}}{{{v_2} - {v_3}}} = \dfrac{{{t_1} - {t_2}}}{{{t_2} - {t_3}}}$
Hence, the correct option is C.

Note
For clarity, we need to note that the use of initial velocity minus final velocity (which was obtained from multiplying both sides by -1) in the actual sense works when only the magnitude of the acceleration is considered.
This is because, since velocity is a vector and time is a scalar, the direction of the change in velocity gets reversed, and this is not counterbalanced by any change in direction of time. However to balance it we make changes in time negative, like saying $10:00 - 10:30 = - 30\min$. The equation in general is just a mathematical equivalence, and not a physical reality.