Question

The acceleration of a particle is given by a = (4t + 9) ms^-2. Then find the displacement of the particle at t = 2 s, if at t = 0 the particle starts from origin with a velocity of 3 ms^-1.

A) 88/3 m
B) 7/9 m
C) 88/9 m
D) 77/18 m​

Answer 1

Topic :- Motion in straight line

$\maltese\:\underline{\sf AnsWer :}\:\maltese$

Given that,the Acceleration (a) of a particle is given by = (4t + 9) m/s². At t = 0 the particle starts from origin with a initial velocity (u) = 3 m/s. We are asked to find the displacement of the particle at t = 2 s.

First of all we need to find the final velocity (v) of the given particle. Acceleration (a) of any particle can be calculated by the given below formula :

$\longrightarrow\:\:\sf a = \dfrac{dv}{dt}$

$\longrightarrow\:\:\sf a.dt = dv$

Now, by integrating both the sides we get :

$\longrightarrow \: \: \displaystyle \sf \int\limits_{0}^{t} a.dt =\int\limits_{v}^{u} dv$

$\longrightarrow \: \: \displaystyle \sf \int\limits_{0}^{t} (4t + 9) \: dt=v - u$

$\longrightarrow \: \:\displaystyle \sf v - u = \int\limits_{0}^{t} (4t + 9) \: dt$

$\longrightarrow \: \:\displaystyle \sf v - 3 = \int\limits_{0}^{t}\dfrac{4 {t}^{1 + 1} }{1 + 1} + \int\limits_{0}^{t} 9t +\int\limits_{0}^{t}C$

$\longrightarrow \: \:\displaystyle \sf v - 3 = \int\limits_{0}^{t}\dfrac{4 {t}^{2} }{2} + \int\limits_{0}^{t} 9t +\int\limits_{0}^{t}C$

$\longrightarrow \: \:\displaystyle \sf v - 3 =\Bigg[\dfrac{4 {t}^{2} }{2}\Bigg]^{t}_{0} + \Bigg[ 9t \Bigg]^{t}_{0} +\Bigg[C\Bigg]^{t}_{0}$

By eliminating "C" because it is constant.

$\longrightarrow \: \:\displaystyle \sf v - 3 =\Bigg[\dfrac{4 {t}^{2} }{2}\Bigg]^{t}_{0} + \Bigg[ 9t \Bigg]^{t}_{0}$

$\longrightarrow \: \:\displaystyle \sf v - 3 =\Bigg[\dfrac{4 {(t)}^{2} }{2} -\dfrac{4 {(0)}^{2} }{2} \Bigg] + \Bigg[ 9(t) - 9(0)\Bigg]$

$\longrightarrow \: \:\displaystyle \sf v - 3 =\Bigg[\dfrac{4 {t}^{2} }{2} -0\Bigg] + \Bigg[ 9t- 0\Bigg]$

$\longrightarrow \: \:\displaystyle \sf v - 3 =\dfrac{4 {t}^{2} }{2} + 9t$

$\longrightarrow \: \:\displaystyle \sf v - 3 = {2t}^{2} + 9t$

$\longrightarrow \: \:\displaystyle \sf v = {2t}^{2} + 9t + 3$

We have find the final velocity (v). Now, let's find the Displacement "x" of the given particle :

$\dashrightarrow\:\:\sf v = \dfrac{dx}{dt}$

$\dashrightarrow\:\:\sf v.dt = dx$

$\dashrightarrow\:\:\sf dx = v.dt$

Integrating both the sides we have :

$\dashrightarrow\: \: \sf\displaystyle \sf \int\limits_{0}^{x} dx =\int\limits_{0}^{2} (2 {t}^{2} + 9t + 3) \: dt$

$\dashrightarrow\: \: \sf x=\int\limits_{0}^{2} 2 {t}^{2} +\int\limits_{0}^{2} 9t + \int\limits_{0}^{2} 3 +\int\limits_{0}^{2} C$

$\dashrightarrow\: \: \sf x=\int\limits_{0}^{2} \dfrac{ {2t}^{2 + 1} }{2 + 1} +\int\limits_{0}^{2} \dfrac{ {9t}^{1 + 1} }{1 + 1} + \int\limits_{0}^{2} 3t + \int\limits_{0}^{2}C$

$\dashrightarrow\: \: \sf x=\int\limits_{0}^{2} \dfrac{ {2t}^{3} }{3} +\int\limits_{0}^{2} \dfrac{ {9t}^{2} }{2} + \int\limits_{0}^{2} 3t + \int\limits_{0}^{2}C$

Eliminating "C" as it is a constant.

$\dashrightarrow\: \: \sf x=\Bigg[ \dfrac{ {2t}^{3} }{3}\Bigg]^2_0 +\Bigg[ \dfrac{ {9t}^{2} }{2} \Bigg]^2_0+ \Bigg[3t \Bigg]^2_0$

$\dashrightarrow\: \: \sf x=\Bigg[ \dfrac{ {2(2)}^{3} }{3} - \dfrac{ {2(0)}^{3} }{3}\Bigg] +\Bigg[ \dfrac{ {9(2)}^{2} }{2} - \dfrac{ {9(0)}^{2} }{2} \Bigg]+ \Bigg[3(2) - 3(0) \Bigg]$

$\dashrightarrow\: \: \sf x=\Bigg[ \dfrac{ {2 \times 8} }{3} - 0\Bigg] +\Bigg[ \dfrac{ 9 \times 4 }{2} - 0 \Bigg]+ \Bigg[6- 0 \Bigg]$

$\dashrightarrow\: \: \sf x =\dfrac{{16} }{3} + \dfrac{36}{2} + 6$

$\dashrightarrow\: \: \sf x =\dfrac{{16} }{3} + 18+ 6$

$\dashrightarrow\: \: \sf x =\dfrac{{16} }{3} + 24$

$\dashrightarrow\: \: \sf x =\dfrac{{16 + 24 \times 3} }{3}$

$\dashrightarrow\: \: \sf x =\dfrac{{16 + 72} }{3}$

$\dashrightarrow\: \: \underline{ \boxed{ \sf x =\dfrac{{88} }{3}}}$

Hence,the displacement of the particle at t = 2 s is x = 88/3. Hence option (A) is the correct answer.