question_answer

The distance covered by an object between times \[{{t}_{1}}\] and \[{{t}_{2}}\] given by the area under the \[\upsilon -t\]graph between \[{{t}_{1}}\] and \[{{t}_{2}}\] Prove this statement for an object moving with negative acceleration and having a positive velocity at time \[{{t}_{1}}\] and a negative velocity at time \[{{t}_{2}}\]

Answer:

                 The velocity-time graph for the given motion of the object is as shown in Fig. Let \[{{\upsilon }_{1}}\] be the velocity at time  \[{{t}_{1}}\] and \[-{{\upsilon }_{2}}\] at time \[{{t}_{2}}\]. Area under \[\upsilon -t\] graph between times \[{{t}_{1}}\] and \[{{t}_{2}}\] \[\text{=Area AA }\!\!'\!\!\text{ B + Area CC }\!\!'\!\!\text{ B}\] \[\text{=}\frac{1}{2}A'B\times AA'+\frac{1}{2}BC'\times CC'\] \[=\frac{1}{2}(t'-{{t}_{1}})({{\upsilon }_{1}})+\frac{1}{2}({{t}_{2}}-t')(-{{\upsilon }_{2}})\] \[=\frac{1}{2}{{\upsilon }_{1}}(t'-{{t}_{1}})-\frac{1}{2}{{\upsilon }_{2}}({{t}_{2}}-t')\]                 Between times \[{{t}_{1}}\] and \[t'\], the acceleration of the object is \[a=\]Slope of \[\upsilon -t\] graph \[AB=\frac{0-{{\upsilon }_{1}}}{t'-{{t}_{1}}}\] \[\therefore \]  \[t'-{{t}_{1}}=-\frac{{{\upsilon }_{1}}}{a}\] Between times \[t'\] and \[{{t}_{2}},\] the acceleration of the objects is \[a=\]slope of \[\upsilon -t\] graph \[BC=\frac{-{{\upsilon }_{2}}-0}{{{t}_{2}}-t'}\] \[\therefore \]  \[{{t}_{2}}-t'=\frac{{{\upsilon }_{2}}}{a}\] Hence area under \[\upsilon -t\] graph between times \[{{t}_{1}}\]and \[{{t}_{2}}\] \[=\frac{1}{2}{{\upsilon }_{1}}\left( -\frac{{{\upsilon }_{1}}}{a} \right)-\frac{1}{2}{{\upsilon }_{2}}\left( -\frac{{{\upsilon }_{2}}}{a} \right)\] \[=\frac{\upsilon _{2}^{2}-\upsilon _{1}^{2}}{2a}=\frac{2as}{2a}=s\] Distance covered by the object between times \[{{t}_{1}}\] and \[{{t}_{2}}\].