Question

the position of a body is given by x= ( 3t⁴ + 4t³ + 2t² + 5t + 1 ) m/s. find velocity at t = 3 seconds. find acceleration at t = 4seconds ​

Answer 1

Answer:

Given displacement x=5+3t+2t

2

Differentiating the above equation with respect to time t we get velocity of the body

v=

dt

dx

=3+4t

Now again differentiating the velocity with respect to time we get acceleration

a=

dt

dv

=4

Explanation:

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Answer 2

Answer:

The position of a body is given by x= ( 3t⁴ + 4t³ + 2t² + 5t + 1 ) m/s.

The velocity at t = 3 seconds will be $245m/s$ and

Acceleration at t = 4seconds ​will be $680m/s^2$

Explanation:

• The velocity can be defined as the rate of change of position with respect to time and

• in the same way, acceleration can be defined as the rate of change of velocity or the second derivative of position with respect to time.

• Let x be the position, v be the velocity and a be the acceleration of an object.

        Then, by definition,

        $\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{d} t}$

        and

        $a=\frac{d v}{d t}$    

• In the question, it is given that,

       $x=3t^{4}+4 t^{3}+2t^2+5 t+1$          

 

• Taking the derivative of x,

      $v=\frac{d (3t^{4}+4 t^{3}+2t^2+5 t+1) }{d t}= 12t^3+12t^2+8t+5$

      put $t=3s$

       $v=(12\times3^3)+(12\times 3^2)+(8\times 3)+5$

           $=245m/s$

• Taking the derivative of v

       $a=\frac{d (12t^3+12t^2+8t+5) }{d t}= 36t^2+ 24t+8$

       put $t= 4s$ to obtain acceleration at t=4s

        $a= (36\times 4^2)+ (24\times 4)+8$

             $= 680m/s^2$