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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,TAccording to the given situation, corresponding v−t, a−t and s−t graphs are as shown below: In the v−t graph, we can see that the slope is constant and positive. In the a−t graph, slope is 0 as the line is parallel to t-axis and is also constant as the slope is not changing. In the s−t graph, we will find that dsdt is negative and the |slope| keeps on decreasing as we move along t-axis.

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