Question

Velocity of a particle is in negative direction with constant acceleration in positive direction. Then match the following.
$\begin{array}{cc}\text{Table-1}& \text{Table-2}\\ \text{(A) Velocity - time graph}& \text{(P) Slope}\to \text{negative}\\ \text{(B) Acceleration - time graph}& \text{(Q) Slope}\to \text{positive}\\ \text{(C) Displacement - time graph}& \text{(R) Slope}\to \text{zero}\\ & \text{(S)}|\text{Slope}|\to \text{increasing}\\ & \text{(T)}|\text{Slope}|\to \text{decreasing}\\ & \text{(U)}|\text{Slope}|\to \text{constant}\end{array}$

• A. $\left(A\right)\to Q,T;\left(B\right)\to Q,S;\left(C\right)\to P,T$
• B. $\left(A\right)\to Q,U;\left(B\right)\to R,U;\left(C\right)\to P,T$
• C. $\left(A\right)\to P,T;\left(B\right)\to R,U;\left(C\right)\to Q,S$
• D. $\left(A\right)\to P,T;\left(B\right)\to Q,U;\left(C\right)\to Q,T$

Answer 1

The correct option is B $\left(A\right)\to Q,U;\left(B\right)\to R,U;\left(C\right)\to P,T$

According 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 $\frac{ds}{dt}$ is negative and the $|slope|$ keeps on decreasing as we move along $t$-axis.