Draw the v-t graph of uniform motion & The area under the v-t graph gives the displacement of the particle in a given time.

True

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False

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Solution

The correct option is A True
Area under -t graph is $=$ area of trapezium
$A r e a = \frac{s u m o f p a r a l l e l s i d e \times h e i g h t}{2}$
$= \frac{(u + v) \times t}{2}$
$= \frac{(u + u + a t) \times t}{2}$
$= u t + \frac{a t^{2}}{2}$
$= s$

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Draw the v-t graph of uniform motion & The area under the v-t graph gives the displacement of the particle in a given time.

The correct option is A TrueArea under -t graph is = area of trapeziumArea=sumofparallelside×height2 =(u+v)×t2 =(u+u+at)×t2 =ut+at22 =s

The correct option is A True
Area under -t graph is $=$ area of trapezium
$A r e a = \frac{s u m o f p a r a l l e l s i d e \times h e i g h t}{2}$
$= \frac{(u + v) \times t}{2}$
$= \frac{(u + u + a t) \times t}{2}$
$= u t + \frac{a t^{2}}{2}$
$= s$