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This is second of our many quizzes on Displacement and distance. Click on this to access more such competitive quizzes on Mechanics.

Displacement and distance Quiz 2

physics

From $v-t$ graph. Distance and displacement for time $0-4\sec $. The graph is semi-circular. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0750c17b-80c1-42fc-917d-f353564c6e4a7345367221773658475.png"/> 

 Calculate the area under $v-t$ to find out distance and displacement. 

  $\text{Area under }v-\text{t graph gives displacement}\text{.} \\ \text{The }v-\text{t curve is in the first quadrant.} \\ \text{So, the value of distance and displacement are equal}\text{.} \\ \text{Distance = Displacement}=\dfrac{\pi {{R}^{2}}}{2}\\ \Rightarrow\text{Distance = Displacement}=\dfrac{\pi }{2}\times 2\times 2 \\ \therefore\text{Distance = Displacement}=2\pi$

Velocity as a function of time is given as $v=\left( 1+2t+3{{t}^{2}} \right)$. Displacement of particle during $1-3\sec $ is

As $\dfrac{ds}{dt}=v$, to calculate displacement of particles we need to use integration technique.

  $\dfrac{ds}{dt}=v \\ \int_{{{S}_{1}}}^{{{S}_{3}}}{ds=\int_{1}^{3}{\left( 1+2t+3{{t}^{2}} \right)dt}} \\ \Rightarrow\int_{{{S}_{1}}}^{{{S}_{3}}}{ds}=\left. t+{{t}^{2}}+{{t}^{3}} \right|_{1}^{3}\\ \Rightarrow\int_{{{S}_{1}}}^{{{S}_{3}}}{ds}=\left[ \left( 3+9+27 \right)-3 \right] \\ \therefore\int_{{{S}_{1}}}^{{{S}_{3}}}{ds}=36\,m$

 In $v-s$ graph i.e velocity displacement graph particle starts with $4m/s$.Initially for $t=1\sec $ displacement of particle is<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7378f6ab-8b8e-4ef2-be00-cb60ab00a7271031653614848467011.png"/>

In the given graph, use $\dfrac{dv}{ds}=$ slope of curve, and use $v=\dfrac{ds}{dt}$ to calculate displacement. 

 $\text{In a given graph,} \\ \dfrac{dv}{ds}=slope=\tan 60{}^\circ =\sqrt{3} \\ \text{Also v}=\sqrt{3}s+\text{4}\,\text{ equation of straight line in }x-y\text{ plane}\\ \dfrac{ds}{dt}=\sqrt{3}s+4 \\ \Rightarrow\int{\dfrac{ds}{\left( \sqrt{3}s+4 \right)}=\int{dt}} \\ \Rightarrow\left. \dfrac{In\left( \sqrt{3}s+4 \right)}{\sqrt{3}} \right|_{0}^{1}=\left. t \right|_{0}^{1} \\ \Rightarrow \left. In\left( \sqrt{3}s+4 \right) \right|_{0}^{1}=\left. \sqrt{3}\times t \right|_{0}^{1} \\ \Rightarrow In\left( \dfrac{\sqrt{3}s+4}{4} \right)=\sqrt{3} \\ \text{Take antilog in both sides}\text{.} \\ \Rightarrow \sqrt{3}s+4=4{{e}^{\sqrt{3}}} \\ \Rightarrow \sqrt{3}s=4\left( {{e}^{\sqrt{3}}}-1 \right) \\ \therefore s=\dfrac{4}{\sqrt{3}}\left( {{e}^{\sqrt{3}}}-1 \right)\,m$

 In $X-t$ curve (distance-time) the value of speed is<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/8e2d0d08-9827-436f-9d1e-18d9a7abdbb33823057741572718290.png"/> 

To calculate speed use $\left( \dfrac{{{x}_{2}}-{{x}_{1}}}{{{t}_{2}}-{{t}_{1}}} \right)$ for time interval $\Delta t$ and for nature of curve draw tangent at multiple instanced. 

 In given graph, $ \\ \left| v \right|=\dfrac{{{x}_{2}}-{{x}_{1}}}{{{t}_{2}}-{{t}_{1}}}\\ $ For any two instance is positive and slope of the curve is decreasing continuous. While we draw tangents at multiple points with increasing time.

 In the $X-t$ graph ${{V}_{1}}$ and ${{V}_{2}}$ are velocities for two lines $\left( 1 \right)$ and $\left( 2 \right)$ $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a55e30dd-20f5-4f99-833d-c55037ed32035063677045367154404.png"/>$ \\ $ The magnitude of velocity is 

Slope of $x-t$ graph gives the value of velocity. Compare ratio to predict the result. 

 In an $x-t$ graph, the slope at any point gives the velocity of the particle. From the graph, it we consider the magnitude of slope of $\left( 1 \right)$ and $\left( 2 \right)$ then $ \\ \left| Slop{{e}_{1}} \right| > {{\left| Slope \right|}_{2}} \\ \therefore \left| {{V}_{1}} \right| > \left| {{V}_{2}} \right|$