Question

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

Answer 1

Through the integration of the acceleration we can get a velocity equation with a constant c. To find the value of c, substitute the value of velocity at ${{t = 1s}}$ then we can get the value of c.
We have to substitute the value of c to find the velocity at $t = 2s$. As per question, displacement can be determined by integrating the velocity.
We have to integrate the velocity equation which we have used in the beginning. Through integrating the velocity equation at ${{t = 1s}}$, we can get the value of c.
Substitute the value of c at ${{t = 2s}}$, we can get displacement s.

Complete step by step answer:
According to question we have, $a = {t^3} - 3{t^2} + 5$
Now we can get the velocity through the integration of acceleration
${{a = }}\dfrac{{dv}}{{dt}} \Rightarrow {{v = }}\int {{{a }}dt}$
$\Rightarrow {{v = }}\int {{{(}}{t^3} - 3{t^2} + 5){{ }}dt}$
Putting the value of $a$ and we get,
$\Rightarrow \int {\left( {{t^3}} \right)} dt - \int {\left( {3{t^2}} \right)} dt + \int {\left( 5 \right)dt}$
On integration we get,
$\Rightarrow {{v}} = \dfrac{{{t^4}}}{4} - \dfrac{{3{t^3}}}{3} + 5t + {{c}}$
At $t = 1s$ the velocity is$6.25{{ }}m{s^{ - 1}}$.
By substituting the values we can get the value of $c$.
$\Rightarrow 6.25 = \dfrac{{{1^4}}}{4} - \dfrac{{3{{(1)}^3}}}{3} + 5(1) + {{c}}$
On simplification we get,
$\Rightarrow {{c = }}6.25 + 1 - 5 - 0.25$
Let us subtracting the decimal values and we get,
$\Rightarrow {{c = 7}} - 5$
On subtract we get,
$\Rightarrow c = 2$
Now, we have to find the velocity at ${{t = 2s}}$.
$\Rightarrow {{v}} = \dfrac{{{2^4}}}{4} - \dfrac{{3{{(2)}^3}}}{3} + 5(2) + 2$
On squaring the values and we get
$\Rightarrow {{v}} = \dfrac{{16}}{4} - \dfrac{{3 \times 8}}{3} + 5(2) + 2$
On cancel the terms and multiply we get,
$\Rightarrow {{v}} = 4 - 8 + 10 + 2$
Let us solve some calculation we get,
$\Rightarrow {{{V}}_{{{t = 2s}}}} = {{ }}8{{ }}m/sec$
The displacement can be determined by integrating the velocity.
$\Rightarrow {{v = }}\dfrac{{ds}}{{dt}}$
Taking integration on both sides we get
$\Rightarrow {{s = }}\int {{{v }}dt}$
Putting the value and we get,
$\Rightarrow {{s = }}\int {\left( {\dfrac{{{t^4}}}{4} - \dfrac{{3{t^3}}}{3} + 5t + 2} \right){{ }}dt}$
On splitting the integral
$\Rightarrow \int {\left( {\dfrac{{{t^4}}}{4}} \right)dt - \int {\left( {\dfrac{{3{t^3}}}{3}} \right)dt + \int {\left( {5t} \right)dt + \int {\left( 2 \right)dt} } } }$
On doing some integration we get
$\Rightarrow {{s =}}\dfrac{{{t^5}}}{{20}} - \dfrac{{{t^4}}}{4} + \dfrac{{5{t^2}}}{2} + 2t + {{c}}$
At $t = 1s$ the displacement is$8.30{{ }}m$. By substituting the values we can get the value of c.
$\Rightarrow {{8}}{{.30 =}}\dfrac{1}{{20}} - \dfrac{1}{4} + \dfrac{5}{2} + 2 + {{c}}$
On doing some simplification we get,
$\Rightarrow {{c = 8}}{{.30}} - 0.05 + 0.25 - 2.5 - 2 = 4$
At $t = 2s$we have to find the displacement.
$\Rightarrow {{s = }}\dfrac{{{2^5}}}{{20}} - \dfrac{{{2^4}}}{4} + \dfrac{{5{{(2)}^2}}}{2} + 2(2) + 4$
On squaring the value and we get
$\Rightarrow {{s = }}\dfrac{{32}}{{20}} - \dfrac{{16}}{4} + \dfrac{{5 \times 4}}{2} + 2(2) + 4$
On doing some simplification we get
$\Rightarrow {{s = }}\dfrac{8}{5} - 4 + 10 + 4 + 4$
On cancel the term we get,
$\Rightarrow {{s}} = 1.6 + 14$
On adding we get
$\Rightarrow s = 15.6{{m}}$
So the displacement and velocity at time $t{{ }} = {{ }}2s$is
$\Rightarrow V = \;8{{ }}m/sec$
$\Rightarrow s = 15.6{{ }}m$

Note: The first derivative of velocity with respect to time is called acceleration.
$a = \dfrac{{dv}}{{dt}}$
$\left( a \right)dt = dv$
$v = \int {\left( a \right)dt}$, where $v =$ velocity, $a =$ acceleration
The first derivative of position with respect to time is called displacement.
$v = \dfrac{{ds}}{{dt}}$
$\left( v \right)dt = ds$
$s = \int {\left( v \right)dt}$, where $v =$ velocity and $s =$ displacement