Question

find acceleration of the moving object where v(f)=3t^2+4t at t=2s​

Answer 1

Answer:

The required acceleration is 16 m/s²

Explanation:

Given, v(t) = 3t² + 4t

• Acceleration is the rate of change of velocity.

We get acceleration by differentiating velocity.

$\boxed{\longrightarrow \sf a=\dfrac{dv}{dt}}$

Differentiating velocity,

$\implies \sf a=\dfrac{dv}{dt} \\\\ \implies \sf a=\dfrac{d}{dt}(3t^2+4t) \\\\ \implies \sf a=\dfrac{d}{dt}(3t^2)+\dfrac{d}{dt}(4t) \\\\ \implies \sf a=6t+4$

∴ The acceleration of the object is given by 6t + 4

Now, we have to find the acceleration at t = 2 s

Put t = 2,

⇒ a = 6t + 4

⇒ a = 6(2) + 4

⇒ a = 12 + 4

⇒ a = 16

So, the required acceleration is 16 m/s²

Answer 2

Answer:

The acceleration of the moving object where v(f)=3t^2+4t at t=2s​ will be $16m/s^2$

Explanation:

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

        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,

        $v = 3t^2+4t$          $t= 2s$                

• Taking the derivative of v,

       $a=\frac{d v}{d t}= 6t+4$

       At time $t= 2s$

       $a= (6\times 2)+4=16 m/s^2$