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A particle converted distance \mathrm{x} meter in \mathrm{t} seconds. The equation is given by- \mathrm{x=0.5 t^{3}+2 t^{2}}
What will be the acceleration at \mathrm{t=2\, \mathrm{sec}}

Option: 1

4 \mathrm{~m} / \mathrm{s}^{2}


Option: 2

8 \mathrm{~m} / \mathrm{s}^{2}


Option: 3

10 \mathrm{~m} / \mathrm{s}^{2}


Option: 4

16 \mathrm{~m} / \mathrm{s}^{2}


Answers (1)

best_answer

Given, \mathrm{x=0.5 t^{3}+2 t^{2}}

To find acceleration firstly try to find velocity
\mathrm{\text { Velocity }(v) =\frac{d x}{d t}=\frac{d}{d t}\left[0.5 t^{3}+2 t^{2}\right]}
                  
                          \mathrm{v =1.5 t^{2}+4 t}

Now, Acceleration is the rate of change of velocity-
\mathrm{a=\frac{d v}{d t} =\frac{d}{d t}\left(1.5 t^{2}+4 t\right)}
   \mathrm{=3.0 t+4}

\mathrm{a=3 t+4 \Rightarrow a =3 \times 2+4}
\mathrm{a =10 \mathrm{~m} / \mathrm{s}^{2}
 

 

,
     

Posted by

vishal kumar

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