Question

find the velocity of the object in interval t=2s to t=3s where a(t) =t^3+3t^2+2 is acceleration of the object in ms^-1​

Answer 1

Answer:

The required velocity is 37.25 m/s  

Explanation:

Given, a(t) = t³ + 3t² + 2

We have to find the velocity of the object in interval t = 2 s to t = 3 s

Integrate acceleration to find velocity of the object.      

$\longmapsto \sf v=\int\limits^{t_2}_{t_1} {a} \, dt$

Integrating,

$\implies \sf v=\int\limits^3_2 {(t^3+3t^2+2)} \, dt \\\\\\ \implies \sf v=\bigg[\dfrac{t^4}{4}+3\bigg(\dfrac{t^3}{3}\bigg)+2t \bigg]^3_2 \\\\\\ \implies \sf v=\bigg[\dfrac{t^4}{4}+t^3+2t \bigg]^3_2 \\\\\\ \implies \sf v=\dfrac{(3)^4-(2)^4}{4} + (3^3-2^3)+2(3-2) \\\\\\ \implies \sf v=\dfrac{81-16}{4}+(27-8)+2(1) \\\\\\ \implies \sf v=\dfrac{65}{4}+19+2 \\\\\\ \implies \sf v=16.25+21 \\\\\\ \implies \sf v=37.25 \ m/s$

Therefore, the required velocity is 37.25 m/s