Question

The acceleration of a particle is given by $a=t^3-3t^2+5$, where $a$ is in $m{s}^{-2}$ and $t$ is in second. At $t=1s$, the displacement and velocity are $8.30m$ and $6.25m{s}^{-1}$, respectively. Calculate the displacement and velocity at $t=2s$.

Answer 1

$a=t^2-3t^2+5$

$v=\int adt+c_1=\frac{t^4}{4}-\frac{3t^3}{3}+5t+C_1$

At $t=1s,v=6.25m{s}^{-1}$

$6.25=\frac{1^4}{4}-1^3+5\times 1+C_1\Rightarrow C_1=2$

$\Rightarrow v=\frac{t^4}{4}-t^3+5t+2$

$x=\int vdt+C_2=\frac{t^5}{20}-\frac{1^4}{4}+\frac{5}{2}(1{)}^2+\times 1+C_2\Rightarrow C_2=4$

$\Rightarrow x=\frac{t^5}{20}-\frac{t^4}{4}+\frac{5}{2}{t}^2+2t+4.$

Now put $t=2s$

$v=\frac{2^4}{4}-2^3+5\times 2+2=8m{s}^{-1}$

$x=\frac{2^5}{20}-\frac{2^4}{4}+\frac{5}{2}(2{)}^2+2\times 2+4=15.6m$