Question

The acceleration of a particle is given by a=t³-3t²+5, where a is in m/s² and t is in seconds. At t=1s, the displacement and velocity are 8.30m and 6.25m/s, respectively. Calculate the displacement and velocity at t=2s.​

Answer 1

Answer:

HOPE IT HELPS YOU!!!

AND IF YOU LIKE THER ANSWER SO PLEASE MARKA ME AS BRAINLIEST!!!

AND ON DIFFERENTIATING THE VELOCITY YOU WILL GET DISPLACNMENT !!!!

Answer 2

GIVEN :-

• Acceleration of a particle is given by a = t³ - 3t² + 5
• At t = 1sec , The displacement and velocity are 8.30m and 6.25m/s

TO FIND :-

• Displacement and velocity at t = 2sec

SOLUTION :-

The relation between velocity and acceleration is given by ,

$\large {\underline {\bold {\boxed { \sf{\bigstar | \: a = \dfrac{dv}{dt} }}}}}$

We have ,

• a = t³ - 3t² + 5

$\implies \sf a = \dfrac{dv}{dt} \\ \\ \implies \sf \: a.dt = dv$

Integrating on both sides ,

$\implies \sf \: \int a \: dt = \int \: dv \\ \\ \implies \sf \: \int \: a \: dt = v \\ \\ \implies \sf \: v = \int \: ( {t}^{3} - 3 {t}^{2} + 5) \: dt$

$\large {\underline{\bold {\boxed { \sf{\bigstar \: \int (u + v )dx = \int u \: dx + \int v \: dx}}}}}$

$\large \underline{\bold {\boxed {\sf {\bigstar \: \int {x}^{n} \: dx = \dfrac{ {x}^{n + 1} }{n + 1} }}}}$

$\implies \sf \: v = \int ( {t}^{3}) \: dt \: - \int \: (3 {t}^{2}) \: dt + \int (5) \: dt \\ \\ \large \underline{\bold {\boxed {\sf {\bigstar \: \int f(x) \: dx = g(x) + c}}}}\\ \\ \implies \sf \: v = \frac{ {t}^{3 + 1} }{(3 + 1)} - \frac{3 {t}^{(2 + 1)} }{2 + 1} + 5t + c_1 \\ \\ \implies \sf \: v = \frac{ {t}^{4} }{4} - \frac{3 {t}^{3} }{3} + 5t + c_1 \\ \\ \implies \sf \: v = \frac{ {t}^{4} }{4} - \frac{ \cancel{3} {t}^{3} }{ \cancel{3}} + 5t + c_1 \\ \\ \implies \: {\sf {\pink{ v= \frac{ {t}^{4} }{4} - {t}^{3} + 5t + c_1}}}$

We are given that ,

• Velocity at t = 1 sec is 6.25 m/s

$\implies \sf \: 6.25 = \frac{(1) {}^{4} }{4} - {(1)}^{3} + 5(1) + c_1 \\ \\ \implies \sf \: 6.25 = \frac{1}{4} - 1 + 5 + c_1 \\ \\ \implies \sf \: 6.25 = \frac{1 + 4( - 1 + 5 + c_1)}{4} \\ \\ \implies \sf \: 6.25 \times 4 = 1 - 4 + 20 - 4c_1 \\ \\ \implies \sf \: 25 = 17 + 4c_1 \\ \\ \implies \sf \: 25 - 17 = 4c_1 \\ \\ \implies \sf \: 8 = 4c_1 \\ \\ \implies \sf \: c_1 = \frac{8}{4} \\ \\ \implies {\bold {\boxed {\pink {\sf{c_1 = 2}}}}}$

Then velocity is ,

$\implies {\bold {\boxed {\pink {\sf{v = \frac{ {t}^{4} }{4} - {t}^{3} + 5t + 2}}}}}$

At t = 2s ,

$\implies \sf \: v = \frac{(2) {}^{4} }{4} - {(2)}^{3} + 5(2) + 2 \\ \\ \implies \sf \: v = \frac{16}{4} - 8 + 10 + 2 \\ \\ \implies \sf \: v = 4 - 8 + 10 + 2 \\ \\ \implies \sf \: v = - 8+ 16 \\ \\ \implies {\underline {\bold {\boxed {\blue {\sf{v = 8 \: ms {}^{ - 1} }}}}}}$

Relation between displacement and velocity is given by ,

$\large \bold {\boxed {\sf {\bigstar | \: v = \frac{dx}{dt} }}}$

$\implies \sf \: v \: dt = dx$

Integrating on both sides ,

$\implies \sf \: \int v \: dt = \int dx \\ \\ \implies \sf x = \int v \: dt \\ \\ \implies \sf \: x = \int \bigg( \frac{ {t}^{4} }{4} - {t}^{3} + 5t + 2 \bigg) \: dt \\ \\ \implies \sf \: x = \int \: \bigg( \frac{ {t}^{4} }{4} \bigg) \: dt - \int \: ( {t}^{3} ) \: dt + \int (5t) \: dt \: + \int \: (2) \: dt \\ \\ \implies \sf \: x = \bigg( \frac{1}{4} \bigg)\int \: ( {t}^{4}) \: dt \: - \frac{ {t}^{(3 + 1)} }{3 + 1} + \frac{5 {t}^{(1 + 1)} }{1 + 1} + 2t + c_2 \\ \\ \implies \sf \: x = \bigg( \frac{1}{4} \bigg) \bigg( \frac{ {t}^{5} }{5} \bigg) - \frac{ {t}^{4} }{4} + \frac{5 {t}^{2} }{2 } + 2t + c_2 \\ \\ \implies \sf \pink{ x = \frac{ {t}^{5} }{20} - \frac{ {t}^{4} }{4} + \frac{5 {t}^{2} }{2} + 2t + c_2}$

We are given that ,

• At t = 1s , x = 8.3 m

$\implies \sf \: 8.3 = \frac{ {(1)}^{5} }{20} - \frac{(1) {}^{4} }{4} + \frac{5(1) {}^{2} }{2} + 2(1) + c_2 \\ \\ \implies \sf \: 8.3 = \frac{1}{20} - \frac{1}{4} + \frac{5}{2} + 2 + c_2 \\ \\ \implies \sf \: 8.3 = 0.05 - 0.25 + 2.5 + 2 + c_2 \\ \\ \implies \sf \: 8.3 = 4.3 + c_2 \\ \\ \implies \sf \: 8.3 - 4.3 = c_2 \\ \\ \implies \: {\bold {\boxed {\pink {\sf{c_2 = 4}}}}}$

Then displacement is ,

$\implies {\bold {\boxed {\pink {\sf{x = \frac{ {t}^{5} }{20} - \frac{ {t}^{4} }{4} + \frac{5 {t}^{2} }{2} + 2t + 4}}}}}$

At t = 2 sec ,

$\implies \sf \: x = \frac{(2) {}^{5} }{20} - \frac{( 2){}^{4} }{4} + \frac{5 {(2)}^{2} }{2} + 2(2) + 4 \\ \\ \implies \sf \: x = \frac{32}{20} - \frac{16}{4} + \frac{20}{2} + 4 + 4 \\ \\ \implies \sf\: x = 1.6 - 4 + 10 + 8 \\ \\ \implies \sf \: x = 19.6 - 4 \\ \\ \implies {\underline {\bold {\boxed {\blue {\sf{x = 15.6 \: m}}}}}}$

∴ The velocity and displacement at t = 2 sec of the given body are 8 m/s and 15.6 m