$v - t$ graph for rectilinear motion is given as shown in the figure. Matched the column on the basic of the $v - t$ graph.

Question

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Answer 1

R.E.F image
A) By definition : Avg. velocity = $\frac{t o t a l d i s p l a c e m e n t}{t o t a l t i m e t a k e n}$
AS $V = \frac{d x}{d t} \Rightarrow △ x = \int x . d t$
So to find displacement, see the area under the from graph:
$\Rightarrow$ (Area under A) $<$ (Area under B) $<$ (Area under C)
$\Rightarrow (△ x)_{A} < (△ x)_{B} < (△ x)_{C}$
$\Rightarrow (A v g . v e l)_{A} < (A v g . v e l)_{B} < (A v g . v e l)_{C} \Rightarrow A n s : 4$
B) REF.Image.
For acceleration check the
slope of v-t graph as
$a = \frac{d v}{d t}$
At $t = t_{1} : m_{A} > m_{B} > M_{C}$
$\Rightarrow a_{A} > a_{B} > a_{C}$
$A n s : 2, 3$
C) REF. Image.
At $t = 0 : m_{C} > m_{B} > m_{A}$
$\Rightarrow a_{C} > a_{B} > a_{A}$
$A n s : 4$
(D) Speed $\equiv$ Magnitude of velocity.
as velocity for all 3 answer has same value
$t = t_{1} \Rightarrow A n s : 1$

Answer 2

R.E.F image

A) By definition : Avg. velocity =

\frac{t o t a l d i s p l a c e m e n t}{t o t a l t i m e t a k e n}

AS

V = \frac{d x}{d t} \Rightarrow △ x = \int x . d t

So to find displacement, see the area under the from graph:

\Rightarrow

(Area under A)

<

(Area under B)

<

(Area under C)

\Rightarrow (△ x)_{A} < (△ x)_{B} < (△ x)_{C}

\Rightarrow (A v g . v e l)_{A} < (A v g . v e l)_{B} < (A v g . v e l)_{C} \Rightarrow A n s : 4

B) REF.Image.

For acceleration check the

slope of v-t graph as

a = \frac{d v}{d t}

At

t = t_{1} : m_{A} > m_{B} > M_{C}

\Rightarrow a_{A} > a_{B} > a_{C}

A n s : 2, 3

C) REF. Image.

At

t = 0 : m_{C} > m_{B} > m_{A}

\Rightarrow a_{C} > a_{B} > a_{A}

A n s : 4

(D) Speed

\equiv

Magnitude of velocity.

as velocity for all 3 answer has same value

t = t_{1} \Rightarrow A n s : 1