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Question

Velocity of a particle is in negative direction with constant acceleration in positive direction. Then match the following.
Table-1Table-2(A) Velocity - time graph(P) Slope negative(B) Acceleration - time graph(Q) Slope positive(C) Displacement - time graph(R) Slope zero(S) |Slope|increasing(T) |Slope|decreasing(U) |Slope|constant

A
(A)Q,T;(B)Q,S;(C)P,T
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B
(A)Q,U;(B)R,U;(C)P,T
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C
(A)P,T;(B)R,U;(C)Q,S
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D
(A)P,T;(B)Q,U;(C)Q,T
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Solution

The correct option is B (A)Q,U;(B)R,U;(C)P,T
According to the given situation, corresponding vt, at and st graphs are as shown below:

In the vt graph, we can see that the slope is constant and positive.
In the at graph, slope is 0 as the line is parallel to t-axis and is also constant as the slope is not changing.
In the st graph, we will find that dsdt is negative and the |slope| keeps on decreasing as we move along t-axis.

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