Question

Find
average

acceleration in 4 sec
3t^2+2t +6​

Answer 1

Given:

• velocity, $\mathsf{v=3t^{2}+2t+6}$
• time, $\mathsf{t=4}$ seconds

To find: average acc.ⁿ in 4 s = ?

Solution: We know that,

$\quad \mathsf{{acc.^{n}}_{average}=\dfrac{dv}{dt}_{at\:time\:t}}$

$\to \mathsf{{acc.^{n}}_{average}=\dfrac{d}{dt}(3t^{2}+2t+6)_{at\:t=4}}$

$\to \mathsf{{acc.^{n}}_{average}=[\:6t+2\:]_{at\:t=4}}$

$\to \boxed{\mathsf{{acc.^{n}}_{average}=26\:unit\:s^{-2}}}$