A point moves with uniform acceleration and V | vedclass

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(B) Let $u_1, u_2, u_3$ and $u_4$ be the velocities at times $t=0, t_1, (t_1+t_2)$ and $(t_1+t_2+t_3)$ respectively. Let $a$ be the uniform accelerati

Vedclass · Accepted Answer

$(V_1 - V_2) : (V_2 - V_3) = (t_1 + t_2) : (t_2 + t_3)$

Vedclass · Answer

(B) Let $u_1, u_2, u_3$ and $u_4$ be the velocities at times $t=0, t_1, (t_1+t_2)$ and $(t_1+t_2+t_3)$ respectively. Let $a$ be the uniform acceleration.The average velocity in an interval is the arithmetic mean of initial and final velocities for that interval:$V_1 = \frac{u_1 + u_2}{2}$,$V_2 = \frac{u_2 + u_3}{2}$,$V_3 = \frac{u_3 + u_4}{2}$Using the equation $v = u + at$:$u_2 = u_1 + at_1$$u_3 = u_2 + at_2 = u_1 + a(t_1 + t_2)$$u_4 = u_3 + at_3 = u_1 + a(t_1 + t_2 + t_3)$Now,calculate the differences:$V_1 - V_2 = \frac{u_1 + u_2}{2} - \frac{u_2 + u_3}{2} = \frac{u_1 - u_3}{2} = \frac{u_1 - (u_1 + a(t_1 + t_2))}{2} = -\frac{a(t_1 + t_2)}{2}$$V_2 - V_3 = \frac{u_2 + u_3}{2} - \frac{u_3 + u_4}{2} = \frac{u_2 - u_4}{2} = \frac{(u_1 + at_1) - (u_1 + a(t_1 + t_2 + t_3))}{2} = -\frac{a(t_2 + t_3)}{2}$Taking the ratio:$\frac{V_1 - V_2}{V_2 - V_3} = \frac{-(a/2)(t_1 + t_2)}{-(a/2)(t_2 + t_3)} = \frac{t_1 + t_2}{t_2 + t_3}$Thus,$(V_1 - V_2) : (V_2 - V_3) = (t_1 + t_2) : (t_2 + t_3)$.

$A$ point moves with uniform acceleration and $V_1, V_2$ and $V_3$ denote the average velocities in the three successive intervals of time $t_1, t_2$ and $t_3$. Which of the following relations is correct?

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