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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line5 Marks
Question
Refer to the graphs in Fig. 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.
✓

Answer

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:
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 $\text{v}=\frac{\text{dx}}{\text{dt}}=0.$ 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 represents negative acceleration. So curve (d) matches with (i).

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