Question

A particle is moving with a velocity of $v=(3+6t+9t^2)\text{cm}/s$. Find out the displacement of the particle in the interval $t=5\text{s}$ to $t=8\text{s}$.

• A. $1280\text{cm}$
• B. $1287\text{cm}$
• C. $1285\text{cm}$
• D. $1280\text{cm}$

Answer 1

The correct option is B $1287\text{cm}$

Given;
$v=(3+6t+9t^2)\text{cm}/s$
As velocity is the rate of change of displacement, we have;
$\Rightarrow \frac{ds}{dt}=(3+6t+9t^2)$
$\Rightarrow ds=(3+6t+9t^2)dt$
$\Rightarrow {\int }_0^{s}ds={\int }_5^8(3+6t+9t^2)dt$
$\Rightarrow s=[3t+3t^2+3t^3{]}_5^8$
$\Rightarrow s=1287\text{cm}$