Question

Given below is a velocity-time graph for an object in motion along a straight line.

With reference to the above given figure, match the Column I (displacement/distance) with Column II (value) and select the correct answer from the codes given below.

Column I		Column II
A	The distance covered by the	1	$8m$ object in time $t=0s$ to $t=2s$.
B	The displacement of the object	2	$+4m$ in time $t=0s$ to $t=2s$.
C	The displacement of the object	3	$4m$ in time $t=0s$ to $t=4s$.
D	The distance of object in time	4	0 $t=0s$ to $t=4s$.

• A. $A-(3),B-(2),C-(1),D-(4)$
• B. $A-(1),B-(2),C-(3),D-(4)$
• C. $A-(3),B-(2),C-(4),D-(1)$
• D. $A-(4),B-(2),C-(1),D-(3)$

Answer 1

Answer: (C)

A. Distance of the object between $t=0s$ to $t=2s$.
$=$ Area under $v-t$ graph
$=$ Area of triangle of base $=2$ and height $=4$
$=\frac{1}{2}\times$ base $\times$ height $=\frac{1}{2}\times 2\times 4=4m$
B. Displacement of the object between $t=0s$ to $t=2s$
$=$ Distance covered in same interval $=+4m$
C. For $t=0s$ to $t=4s$,
Displacement $=$ Area under $v-t$ curve $(t=0s$ to
$t=2s)+$ Area under $v-t$ curve $(t=2s$ to $t=4s)$
$=\frac{1}{2}\times 2\times 4+\frac{1}{2}\times 2\times (-4)=4m+(-4m)=0$
D. For $t=0s$ to $t=4s$,
Distance covered $=$ Area under $v-t$ curve considering all areas as positive.
$=$ Area under $v-t$ curve $(t=0s$ to $t=2s)$
$+$ Area under $v-t$ curve $(t=2s$ to $t=4s)$ $=4m+4m=8m$
Hence, $A\to 3,B\to 2,C\to 4$ and $D\to 1$.