Question

Acceleration of the car given a=t^2+2t+1.Find velocity after 6seconds​

Answer 1

Given :

Acceleration of the car given

a=t²+2t+1

To find :

Velocity of car after 6 sec .

Solution :

We know that

Acceleration= chane in velocity / time =dv/dt

We have ,a=t²+2t+1

$\sf\implies\dfrac{dv}{dt}=t^2+2t+1$

$\sf\implies\:dv=(t^2+2t+1)dt$

On Integrating both sides

$\sf\int\:dv=\int(t^2+2t+1)dt$

Apply limits to the Integration

$\sf\int\limits_{0}^v=\int\limits_{0}^6(t^2+2t+2)dt$

$\sf\:V=[\dfrac{t^3}{3}+\dfrac{t^2}{2}+t]^6_0$

$\sf\:V=\dfrac{6^3}{3}+\dfrac{6^2}{2}+6$

$\sf\:V=96ms{}^{-1}$

Therefore, Velocity of car after 6 sec is 96m/s.

$\rule{200}2$

Formulas of Integration :

$\sf\int\:x^ndx=\dfrac{x{}^{n+1}}{n+1}$

$\sf\int\dfrac{1}{x}=\log\:x$

Answer 2

Answer:

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