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Question

(A) Let the total distance be $2d$. For the first half distance $d$, the speed is $v_1 = 3 \, m \, s^{-1}$. The time taken is $t_1 = d / v_1 = d / 3$.

Vedclass · Accepted Answer

$4.0 \, m \, s^{-1}$

Vedclass · Answer

(A) Let the total distance be $2d$. For the first half distance $d$, the speed is $v_1 = 3 \, m \, s^{-1}$. The time taken is $t_1 = d / v_1 = d / 3$. For the second half distance $d$, it is covered in two equal time intervals $t_2$ and $t_2$ (total time $2t_2$) with speeds $v_2 = 4.5 \, m \, s^{-1}$ and $v_3 = 7.5 \, m \, s^{-1}$. The distance covered in this half is $d = v_2 t_2 + v_3 t_2 = (4.5 + 7.5) t_2 = 12 t_2$. So, $t_2 = d / 12$. The total time taken for the second half is $T_2 = 2 t_2 = 2(d / 12) = d / 6$. The total time for the whole journey is $T = t_1 + T_2 = d / 3 + d / 6 = (2d + d) / 6 = 3d / 6 = d / 2$. The average speed is $v_{avg} = 	ext{Total Distance} / 	ext{Total Time} = 2d / (d / 2) = 4 \, m \, s^{-1}$.

Column $I$	Column $II$
$(A)$ Distance travelled in $3\,s$	$(p)$ $-20$ units
$(B)$ Displacement in $1\,s$	$(q)$ $15$ units
$(C)$ Initial acceleration	$(r)$ $25$ units
$(D)$ Velocity at $4\,s$	$(s)$ $-10$ units

Similar Questions

$A$ bogie of a uniformly moving train is suddenly detached from the train and stops after covering some distance. What is the relation between the distance covered by the bogie and the distance covered by the train in the same time?

In the $s-t$ equation $(s=10+20t-5t^2)$,match the following columns.

Column $I$ Column $II$

$(A)$ Distance travelled in $3\,s$ $(p)$ $-20$ units

$(B)$ Displacement in $1\,s$ $(q)$ $15$ units

$(C)$ Initial acceleration $(r)$ $25$ units

$(D)$ Velocity at $4\,s$ $(s)$ $-10$ units

Which of the following curves possibly represent one-dimensional motion of a particle? Choose the correct answer from the options given below.

The figure shows the velocity $(v)$ of a particle plotted against time $(t)$.

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