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(C) For the first part of the journey: $u = 0, a = 4 \,m/s^2$, time $= t_1$. The velocity attained at the end of time $t_1$ is $v_1 = u + at_1 = 4t_1$

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(C) For the first part of the journey: $u = 0, a = 4 \,m/s^2$, time $= t_1$. The velocity attained at the end of time $t_1$ is $v_1 = u + at_1 = 4t_1$.The displacement in the first $t_1$ seconds is $s_1 = \frac{1}{2}at_1^2 = \frac{1}{2} 	imes 4 	imes t_1^2 = 2t_1^2$.For the second part of the journey: initial velocity $v_1 = 4t_1$, time $= t_2$, and acceleration $= 0$.The displacement in the next $t_2$ seconds is $s_2 = v_1 t_2 = 4t_1 t_2$.For the third part of the journey: initial velocity $v_1 = 4t_1$, final velocity $= 0$, time interval $= t_1$.The displacement in the third part is $s_3 = \frac{v_1 + v_f}{2} 	imes t_1 = \frac{4t_1 + 0}{2} 	imes t_1 = 2t_1^2$.Total time $T = t_1 + t_2 + t_1 = 2t_1 + t_2 = 10 \,s$, so $t_2 = 10 - 2t_1$.Average velocity $v_{	ext{avg}} = \frac{	ext{Total displacement}}{	ext{Total time}} = \frac{s_1 + s_2 + s_3}{10} = 5.1$.Substituting the values: $5.1 = \frac{2t_1^2 + 4t_1 t_2 + 2t_1^2}{10} = \frac{4t_1^2 + 4t_1(10 - 2t_1)}{10}$.$51 = 4t_1^2 + 40t_1 - 8t_1^2$.$4t_1^2 - 40t_1 + 51 = 0$.Solving the quadratic equation: $t_1 = \frac{40 \pm \sqrt{1600 - 4(4)(51)}}{2(4)} = \frac{40 \pm \sqrt{1600 - 816}}{8} = \frac{40 \pm \sqrt{784}}{8} = \frac{40 \pm 28}{8}$.$t_1 = \frac{68}{8} = 8.5 \,s$ (not possible) or $t_1 = \frac{12}{8} = 1.5 \,s$.Therefore, $t_1 = 1.5 \,s$.

Column-$I$	Column-$II$
$(1)$ Velocity-time relation	$(a)$ $v = v_0 + at$
$(2)$ Velocity-displacement relation	$(b)$ $S = v_0t + \frac{1}{2}at^2$
	$(c)$ $v^2 = v_0^2 + 2as$

Similar Questions

Match the relations in Column-$I$ with the equations in Column-$II$ for uniformly accelerated motion.

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$(1)$ Velocity-time relation $(a)$ $v = v_0 + at$

$(2)$ Velocity-displacement relation $(b)$ $S = v_0t + \frac{1}{2}at^2$

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