Question

a=(t²+t+1)m/s² ,x=?,t=1sec​

Answer 1

Question :

Acceleration of a particle is given by, a = (t² + t + 1) m/s. Find the displacement of the particle at t = 1 s.

Solution :

$\textsf{To find the displacement of the particle,} \\ \textsf{first we have to find the velocity of the particle.}$

To find the velocity of the particle :

$\textsf{We know the formula for velocity of a particle .i.e, } \: \sf{v} = \displaystyle \int \sf{adt} \\ \\ \textsf{So, substituting the values in it and by integrating it, we get :-}$

$:\implies \sf{v} = \displaystyle \int \sf{adt}$

$:\implies \sf{v} = \displaystyle \int \sf{(t^{2} + t + 1)dt}$

$:\implies \sf{v} = \displaystyle \int \sf{t^{2}dt} + \displaystyle \int \sf{tdt} + \displaystyle \int \sf{1dt}$

$\textsf{Now, by using the power rule of integration, we get :-} \\ \\ \underline{\sf{Power\:rule\: integration :}} \\ \\ \displaystyle \sf \int x^{n}dx = \sf\dfrac{x^{(n + 1)}}{n + 1} + c\:, [Where, n \ne -1] \\ \\ \underline{\sf{Constant\:rule\: integration :}} \\ \\ \displaystyle \sf \int n dx = n\cdot x + c$

$:\implies \sf{v} = \sf\dfrac{t^{(2 + 1)}}{2 + 1} + \sf\dfrac{t^{(1 + 1)}}{1 + 1} + \displaystyle \int \sf{1dt}$

$:\implies \sf{v} = \sf\dfrac{t^{3}}{3} + \sf\dfrac{t^{2}}{2} + t + c$

$:\implies \sf{v} = \sf\dfrac{t^{3}}{3} + \sf\dfrac{t^{2}}{2} + t + c$

$\therefore \sf{v} = \sf\dfrac{t^{3}}{3} + \sf\dfrac{t^{2}}{2} + t$

To find the displacement of the particle :

$\textsf{We know the formula for acceleration of a particle .i.e, } \: \sf{x} = \displaystyle \int \sf{vdt} \\ \\ \textsf{So, substituting the values in it and by integrating it, we get :-}$

$:\implies \sf{x} = \displaystyle \int \sf{vdt}$

$:\implies \sf{x} = \displaystyle \int \sf{\left(\dfrac{t^{3}}{3} + \dfrac{t^{2}}{2} + t\right)dt}$

$:\implies \sf{x} = \displaystyle \int \sf{\dfrac{t^{3}}{3}dt} + \displaystyle \int \sf{\dfrac{t^{2}}{2}dt} + \displaystyle \int \sf{tdt}$

$:\implies \sf{x} = \dfrac{1}{3} \displaystyle \int \sf{t^{3}dt} + \dfrac{1}{2} \displaystyle \int \sf{t^{2}dt} + \displaystyle \int \sf{tdt}$

$\textsf{Now, by using the power rule of integration, we get :-} \\ \\ \underline{\sf{Power\:rule\: integration :}} \\ \\ \displaystyle \sf \int x^{n}dx = \sf\dfrac{x^{(n + 1)}}{n + 1} + c\:, [Where, n \ne -1] \\ \\ \underline{\sf{Constant\:rule\: integration :}} \\ \\ \displaystyle \sf \int n dx = n\cdot x + c$

$:\implies \sf{x = \dfrac{1}{3} \cdot \dfrac{t^{(3 + 1)}}{3 + 1} + \dfrac{1}{2} \cdot \dfrac{t^{(2 + 1)}}{2 + 1} + \dfrac{t^{(1 + 1)}}{1 + 1} + c}$

$:\implies \sf{x = \dfrac{1}{3} \cdot \dfrac{t^{4}}{4} + \dfrac{1}{2} \cdot \dfrac{t^{3}}{3} + \dfrac{t^{2}}{2} + c}$

$:\implies \sf{x = \dfrac{t^{4}}{12} + \dfrac{t^{3}}{6} + \dfrac{t^{2}}{2} + c}$

$\therefore \sf{x = \dfrac{t^{4}}{12} + \dfrac{t^{3}}{6} + \dfrac{t^{2}}{2}}$

Now, to find the velocity of the particle at t = 1 s.

$:\implies \sf{x = \dfrac{t^{4}}{12} + \dfrac{t^{3}}{6} + \dfrac{t^{2}}{2}}$

$:\implies \sf{x_{(t = 1 s)} = \dfrac{1^{4}}{12} + \dfrac{1^{3}}{6} + \dfrac{1^{2}}{2}}$

$:\implies \sf{x_{(t = 1 s)} = \dfrac{1}{12} + \dfrac{1}{6} + \dfrac{1}{2}}$

$:\implies \sf{x_{(t = 1 s)} = \dfrac{1 + 2 + 6}{12}}$

$:\implies \sf{x_{(t = 1 s)} = \dfrac{7}{12}}$

$\therefore \sf{x_{(t = 1 s)} = \dfrac{7}{12}}$

$\textsf{Hence, the velocity of the particle at t = 1 s is} \: \sf\dfrac{7}{12} \:m$