Refer to the graph in figure. Match the following
Graph
Characteristics
$(A)$
$(i)$ has $v > 0$ and $a < 0$ throughout
$(B)$
$(ii)$ has $x > 0,$ throughout and has a point with $v = 0$ and a point with $a = 0$
$(C)$
$(iii)$ has a point with zero displacement for $t > 0$
$(D)$
$(iv)$ has $v < 0$ and $a > 0$
Refer to the graph in figure. Match the following
| Graph | Characteristics |
| $(A)$ | $(i)$ has $v > 0$ and $a < 0$ throughout |
| $(B)$ | $(ii)$ has $x > 0,$ throughout and has a point with $v = 0$ and a point with $a = 0$ |
| $(C)$ | $(iii)$ has a point with zero displacement for $t > 0$ |
| $(D)$ | $(iv)$ has $v < 0$ and $a > 0$ |
We have to analyse slope of each curve i.e., $\frac{d x}{d t}$ For peak points $\frac{d x}{d t}$ will be zero as $x$ is maximum at peak points.
For graph $(a)$, there is a point $(B)$ for which displacement is zero. So, a matches with $(iii)$
In graph (b), $x$ is positive $(>0)$ throughout and has a point $\mathrm{B}_{1}$ with $\mathrm{V}=\frac{d x}{d t}=0$. As at point of curvature changes $a=0$, So $b$ matches with $(ii)$
In graph $(c)$, slope $\mathrm{V}=\frac{d x}{d t}$ is negative hence, velocity will be negative. So, it matches with (iv)
In graph $(d)$, as slope $\mathrm{V}=\frac{d x}{d t}$ is positive hence, $\mathrm{V}>0$
Hence, $d$ matches with $(i)$
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- [JEE MAIN 2019]