Question

an object starts linear motion with a velocity 'u' and under uniform acceleration 'a' . it acquires a velocity 'v' in time 't'. draw velicit time graph . from this graph obtain the (a) first equation of motion v= u+at for velocity -time relation . (b) third equation of motion v^-u^=2as for position -velocity relation

Answer 1

Explanation:

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Answer 2

Answer:

From the velocity- time graph (attached below),

• For the first equation of motion, velocity-time realtion:

        We get,  v =EC= OD, u = OA, t = AB=OE

            Acceleration(a) = slope of AC

                                      $=\frac{BC}{AB}=\frac{AD}{OE}$

                                   $a=\frac{OD-OA}{OE}= \frac{v-u}{t}$

                       ∴         $at=v-u$

               we get,      $v=u+at$                            .............(1)

• For the third equation of motion, position-velocity relation:

         S (displacement) = Area of trapezium(OACE)

                                       $=\frac{1}{2} (OA+EC)(OE)$

                                       $=\frac{1}{2} (u+v) (t)$    

            Put the value of 't' equation (1) ,                    

                                       $S=\frac{1}{2} (v+u) (\frac{v-u}{a} )$

                                      $S=\frac{v^{2}-u^{2} } {2a}$

         ∴                       $v^{2} -u^{2} = 2aS$