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

Hint: 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