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	$8 \,m$ object in time $t=0 \,s$ to $t=2\, s$.
B	The displacement of the object	2	$+4\, m$ in time $t=0\, s$ to $t=2 \,s$.
C	The displacement of the object	3	$4\, m$ in time $t=0\, s$ to $t=4 \,s$.
D	The distance of object in time	4	0 $t=0\, s$ to $t=4 \,s$.

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