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This is one of our many quizzes on Position, velocity, and acceleration. Click on this to access more such competitive quizzes on Number System

Position, velocity, and acceleration Quiz 2

mathematics

 Find the velocity of a particle whose position is given by  $S\left( t \right)=\left( 21{{t}^{2}}+t-4 \right)$ with respect to $t$ (time) at $t=23$$\text{seconds}$.

 Velocity is the rate of change of position of an object with respect to time. 

 $  S\left( t \right)=\left( 21{{t}^{2}}+t-4 \right) \\  \Rightarrow S'\left( t \right)=\dfrac{dS}{dt} \\  \Rightarrow 42t+1 \\  \text{At }\!\!~\!\!\text{ }t=23\text{seconds} \\  \Rightarrow v=S'\left( t \right)=\left( 42\times 23+1 \right)\text{m/s} \\  \Rightarrow 967\text{m/s} \\ $

 Find the acceleration of an object whose position is given by $S\left( t \right)=\left( 5{{t}^{3}}+21{{t}^{2}}+19t+1 \right)$ with respect to time at $t=5$ second.

 Derive the equation $S\left( t \right)$ with respect to time to get velocity and then derive further to get the acceleration $S''\left( t \right)$.

 $ S\left( t \right)=\left( 5{{t}^{3}}+21{{t}^{2}}+19t+1 \right) \\  \Rightarrow S\left( t \right)=15{{t}^{2}}+42t+49 \\  \Rightarrow S''\left( t \right)=30t+42 \\  \text{At }\!\!~\!\!\text{  }t=5\text{ second} \\  \Rightarrow a=S''\left( t \right)=\left( 5\times 30+42 \right) \\  \Rightarrow 150+42 \\  \Rightarrow 192\text{m/}{{\text{s}}^{\text{2}}} \\ $

 Find the instantaneous velocity and acceleration of an object whose position time relation is given by  $S\left( t \right)=\left( \dfrac{3}{2}{{t}^{2}}+\dfrac{2}{19}t+60 \right)$ at $t=23$ seconds.

 Velocity is the derivative of position of an object while acceleration is the derivative of velocity with respect to time respectively.

 $ S\left( t \right)=\left( \dfrac{3}{2}{{t}^{2}}+\dfrac{2}{19}t+60 \right) \\  \Rightarrow S'\left( t \right)=\left( 2\times \dfrac{3}{2}t+\dfrac{2}{19} \right) \\  \text{At }\!\!~\!\!\text{  }t=23\text{ seconds} \\  \Rightarrow S'\left( t \right)=23\times 3+\dfrac{2}{19} \\  \Rightarrow 69+\dfrac{2}{19} \\  \Rightarrow \dfrac{1311+2}{19} \\  \Rightarrow \dfrac{1313}{19}\text{m/s} \\  \Rightarrow \text{S}''\left( \text{t} \right)=3\text{m/}{{\text{s}}^{\text{2}}} \\  \text{At }t=23\text{ seconds} \\  \Rightarrow \text{S}''\left( \text{t} \right)=\text{3m/}{{\text{s}}^{\text{2}}} \\ $

 Find the instantaneous acceleration of a object whose position is given by $S\left( t \right)=\left( {{t}^{4}}-26{{t}^{3}}+t+5 \right)$ at $t=7$ seconds and $t=9$ seconds. 

 Simply get the derivative of $S\left( t \right)$ for velocity and for acceleration get the derivative of velocity $S'\left( t \right)$ and put the values of time.

 $ S\left( t \right)=\left( {{t}^{4}}-26{{t}^{3}}+t+5 \right) \\  \Rightarrow S'\left( t \right)=\left( 4{{t}^{3}}-78{{t}^{3}}+1 \right) \\  \Rightarrow S''\left( t \right)=\left( 12{{t}^{2}}-156t \right) \\  \text{At }\!\!~\!\!\text{  }t=7\text{ seconds} \\  \Rightarrow S''\left( t \right)=12\times 49-156\times 7 \\  \Rightarrow 58-1092 \\  \Rightarrow \left( -504 \right)\text{m/}{{\text{s}}^{\text{2}}} \\  \text{At }\!\!~\!\!\text{  }t=9\text{ seconds} \\  \Rightarrow S''\left( t \right)=12\times 81-156\times 9 \\  \Rightarrow \left( -432 \right)\text{m/}{{\text{s}}^{\text{2}}} \\ $

 What is the instantaneous velocity of a particle whose position is given by $S\left( t \right)=\left( 2{{t}^{5}}-{{t}^{4}}+19 \right)$ at $t=3$ and $t=9$.

 Velocity changes position so find the derivative of $S\left( t \right)$ for velocity and acceleration is change in velocity so find $S''\left( t \right)$ for acceleration and put the values of time.

 $  S\left( t \right)=\left( 2{{t}^{5}}-{{t}^{4}}+19 \right) \\  \Rightarrow S'\left( t \right)=\left( 10{{t}^{4}}-4{{t}^{3}} \right) \\  \text{At }t=9 \\  \Rightarrow 10{{\left( 9 \right)}^{4}}-4{{\left( 9 \right)}^{3}} \\  \Rightarrow 65610-2919 \\  \Rightarrow 62694\text{m/s} \\  \text{At }t=3 \\  \Rightarrow 10{{\left( 3 \right)}^{4}}-4{{\left( 3 \right)}^{3}} \\  \Rightarrow 810-108 \\  \Rightarrow 702\text{m/s} \\ $