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प्रश्न
Refer to the graphs in figure. Match the following.
| Graph | Characteristic |
(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. |
उत्तर
| Graph | Characteristic |
| (a) |
(iii) has a point with zero displacement for t > 0. |
| (b) |
(ii) has x > 0 throughout and has a point with v = 0 and a point with a = 0. |
| (c) |
(iv) has v < 0 and a > 0. |
| (d) |
(i) has v > 0 and a < 0 throughout. |
Explanation:
Let us pick graphs one by one.
In graph (a),
There is a point (B) on the curve for which displacement is zero. So curve, (a) matches with (iii).
In graph (b),
In this graph, x is positive (> 0) throughout and at point B the highest point of curve the slope of curve is zero. It means at this point v = dx/dt = 0 . Also at point C the dt
curvature changes, it means at this point the acceleration of the particle should be zero or a = 0, So curve (b) matches with (ii).
In graph (c),
In this graph the slope is always negative, hence velocity will be negative or v < 0. Also, x-t graph opens up, it represents positive acceleration. So curve (c) matches with (iv).
In graph (d),
In this graph the slope is always positive, hence velocity will be positive or v > 0. Also, x-t graph opens down, it representing negative acceleration. So curve (d) matches with (i).
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