Physics-
General
Easy

Question

A graph of moving body with constant acceleration is given in the figure. What is the velocity aftertime t ?

  1. 0 text end text A plus fraction numerator D C over denominator B C end fraction times 0 E    
  2. 0 text end text A plus fraction numerator D C over denominator B C end fraction times D E    
  3. A B plus fraction numerator B C over denominator D C end fraction minus 0 E    
  4. 0 text end text A plus fraction numerator D C over denominator B C end fraction times A D    

The correct answer is: 0 text end text A plus fraction numerator D C over denominator B C end fraction times 0 E

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vertical line a with not stretchy bar on top vertical line equals vertical line b with not stretchy bar on top vertical line equals vertical line c with not stretchy bar on top vertical line equals vertical line a with not stretchy bar on top plus b with not stretchy bar on top plus c with not stretchy bar on top vertical line equals 1  and a with not stretchy bar on top perpendicular b with not stretchy bar on top left parenthesis c with not stretchy bar on top comma a with not stretchy bar on top right parenthesis equals alpha comma left parenthesis c with not stretchy bar on top comma b with not stretchy bar on top right parenthesis equals beta comma t h e n space cos alpha plus cos space beta equals

The given vectors are a with rightwards arrow on top comma b with rightwards arrow on top a n d c with rightwards arrow on top
The given conditions are as follows:
open vertical bar a with rightwards arrow on top close vertical bar equals open vertical bar b with rightwards arrow on top close vertical bar equals open vertical bar c with rightwards arrow on top close vertical bar equals 1
open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top close vertical bar equals 1
a with rightwards arrow on top perpendicular b with rightwards arrow on top
The angle between the vectors c with rightwards arrow on top space a n d space a with rightwards arrow on top is α.
The angle between the vectors c with rightwards arrow on top a n d b with rightwards arrow on top is β.
We have to find the value of cosα + cosβ.
Let's take dot product of pair of vectors.
a with rightwards arrow on top. b with rightwards arrow on top equals vertical line a with rightwards arrow on top vertical line open vertical bar b with rightwards arrow on top close vertical bar cos 90 to the power of 0 space
a with rightwards harpoon with barb upwards on top. b with rightwards arrow on top space equals space 0
b with rightwards arrow on top. c with rightwards arrow on top equals open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar cos beta
space space space space space space space space equals left parenthesis 1 right parenthesis left parenthesis 1 right parenthesis cos beta
space space space space space space space space equals cos beta
c with rightwards arrow on top. a with rightwards arrow on top equals open vertical bar c with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos alpha
space space space space space space space equals left parenthesis 1 right parenthesis left parenthesis 1 right parenthesis cos alpha
space space space space space space space equals cos alpha
Now, we will use the given condition to solve further.
open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top close vertical bar equals square root of left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top right parenthesis. left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top right parenthesis end root
space space space space space space space space space space space space space space 1 space equals square root of left parenthesis open vertical bar a with rightwards arrow on top close vertical bar squared plus space open vertical bar b with rightwards arrow on top close vertical bar squared plus open vertical bar c with rightwards arrow on top close vertical bar squared plus space 2 left parenthesis a with rightwards arrow on top. b with rightwards arrow on top plus b with rightwards arrow on top. c with rightwards arrow on top plus c with rightwards arrow on top. a with rightwards arrow on top right parenthesis end root
Substituting all the values we get,
1 equals square root of 1 plus 1 space plus space 1 space plus 2 cos alpha plus 2 cos beta end root
1 space equals square root of 3 space plus space 2 cos alpha space plus space 2 cos beta end root
S q u a r i n g space b o t h space t h e space s i d e s
1 space equals space 3 space plus space 2 cos alpha space plus space 2 cos beta
minus 2 space equals space 2 cos alpha space plus space 2 cos beta
R e a r r a n g i n g
2 cos alpha space plus space 2 cos beta space equals space minus 2
cos alpha space plus space cos beta space equals space minus 1
So, the answer is -1.

vertical line a with not stretchy bar on top vertical line equals vertical line b with not stretchy bar on top vertical line equals vertical line c with not stretchy bar on top vertical line equals vertical line a with not stretchy bar on top plus b with not stretchy bar on top plus c with not stretchy bar on top vertical line equals 1  and a with not stretchy bar on top perpendicular b with not stretchy bar on top left parenthesis c with not stretchy bar on top comma a with not stretchy bar on top right parenthesis equals alpha comma left parenthesis c with not stretchy bar on top comma b with not stretchy bar on top right parenthesis equals beta comma t h e n space cos alpha plus cos space beta equals

Maths-General
The given vectors are a with rightwards arrow on top comma b with rightwards arrow on top a n d c with rightwards arrow on top
The given conditions are as follows:
open vertical bar a with rightwards arrow on top close vertical bar equals open vertical bar b with rightwards arrow on top close vertical bar equals open vertical bar c with rightwards arrow on top close vertical bar equals 1
open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top close vertical bar equals 1
a with rightwards arrow on top perpendicular b with rightwards arrow on top
The angle between the vectors c with rightwards arrow on top space a n d space a with rightwards arrow on top is α.
The angle between the vectors c with rightwards arrow on top a n d b with rightwards arrow on top is β.
We have to find the value of cosα + cosβ.
Let's take dot product of pair of vectors.
a with rightwards arrow on top. b with rightwards arrow on top equals vertical line a with rightwards arrow on top vertical line open vertical bar b with rightwards arrow on top close vertical bar cos 90 to the power of 0 space
a with rightwards harpoon with barb upwards on top. b with rightwards arrow on top space equals space 0
b with rightwards arrow on top. c with rightwards arrow on top equals open vertical bar b with rightwards arrow on top close vertical bar open vertical bar c with rightwards arrow on top close vertical bar cos beta
space space space space space space space space equals left parenthesis 1 right parenthesis left parenthesis 1 right parenthesis cos beta
space space space space space space space space equals cos beta
c with rightwards arrow on top. a with rightwards arrow on top equals open vertical bar c with rightwards arrow on top close vertical bar open vertical bar a with rightwards arrow on top close vertical bar cos alpha
space space space space space space space equals left parenthesis 1 right parenthesis left parenthesis 1 right parenthesis cos alpha
space space space space space space space equals cos alpha
Now, we will use the given condition to solve further.
open vertical bar a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top close vertical bar equals square root of left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top right parenthesis. left parenthesis a with rightwards arrow on top plus b with rightwards arrow on top plus c with rightwards arrow on top right parenthesis end root
space space space space space space space space space space space space space space 1 space equals square root of left parenthesis open vertical bar a with rightwards arrow on top close vertical bar squared plus space open vertical bar b with rightwards arrow on top close vertical bar squared plus open vertical bar c with rightwards arrow on top close vertical bar squared plus space 2 left parenthesis a with rightwards arrow on top. b with rightwards arrow on top plus b with rightwards arrow on top. c with rightwards arrow on top plus c with rightwards arrow on top. a with rightwards arrow on top right parenthesis end root
Substituting all the values we get,
1 equals square root of 1 plus 1 space plus space 1 space plus 2 cos alpha plus 2 cos beta end root
1 space equals square root of 3 space plus space 2 cos alpha space plus space 2 cos beta end root
S q u a r i n g space b o t h space t h e space s i d e s
1 space equals space 3 space plus space 2 cos alpha space plus space 2 cos beta
minus 2 space equals space 2 cos alpha space plus space 2 cos beta
R e a r r a n g i n g
2 cos alpha space plus space 2 cos beta space equals space minus 2
cos alpha space plus space cos beta space equals space minus 1
So, the answer is -1.
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