Q12 of 26 Page 16

Refer to the graphs in Fig 3.1. 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.


(a) - (iii)


(b) – (ii)


(c) – (iv)


(d) – (i)


(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.



x > 0 throughout in this graph and since the slope in changing. Thus, there is a point of a = 0 and a point of v = 0.


(c) – (iv) has v < 0 and a > 0.



Since, the slope is negative throughout, v < 0 and since the graph is concave upwards (second derivative is greater than zero), a > 0.


(d) – (i) has v > 0 and a < 0 throughout.



The slope is positive throughout the graph, thus, v > 0. The graph is concave downwards (second derivative of x is negative), thus, acceleration is negative.


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