Question

The acceleration of a particle in ms-² is given
by a = 4t³ + t + 3, where t is in second.
If the particle starts with velocity v=2ms-¹
at t=0 then find velocity at the end of 2s..



gyus pls answer it fast​

Answer 1

Given

• An acceleration of a particle in m/s-²
• a = 4t³ + t + 3,

• velocity of particle in starting is 2m/s at t=0
• u=3m/s

To Find

• velocity at t =,2

Solution

The value of acceleration at t=2

We know that

a = dv/dt

a = 4t³ + t + 3,

=>dv=adt

V= 3t³/3+2t²/2+2t

at t =0 and u=2

At t=2

V= t³+t²+2t+2

V=8+4+4+2

V=18

Hence, velocity be 18m/s at time 2

Answer 2

Answer:

$\bigstar$Given:

$\red{\implies \rm{a = 4t^3 + t + 3}}$

⇒ Particle starts with velocity v=2m/s  at t=0

$\bigstar$To Find:

⇒ velocity at the end of 2s

$\bigstar$Solution:

$\implies \rm{a = 4t^3 + t + 3}$

• t is in second

Mathematically, acceleration is given as,

$\blue{\boxed{\rm a = \dfrac{dv}{dt}}}$

$\green{\implies \rm{\dfrac{dv}{dt} = 4t^3 + t + 3}}$

Now to get $\rm v$

We should integrate the above equation,

→ (dv) with limits 2 to v

→ (4t³ + t + 3) with respect to t and with limits 0 to 2

INTEGRATION PROCESS

$\implies \rm dv = (4t^3 + t + 3})\,dt$

$\implies {\int\limits^v_2 { \rm{dv} }= \int\limits^2_0 ({4t^3 + t + 3})dt}}$

$\implies \rm {[v]_2^v= [t^4 + \dfrac{t^2}{2} + 3t}]_0^2$

Using the integration formula,

$\purple{\boxed{\boxed{\rm \int\ {x^n} \, dx = \dfrac{x^{n+1}}{n+1}}}}$

Now apply the integral limits to the equation obtained after integration,

i.e., Upper limit - Lower limit

$\implies \rm {v-2 = (2)^4 + \dfrac{(2)^2}{2} + 3(2) - 0}$

$\implies \rm{v-2 = 16 + 2 + 6 }$

$\implies \rm{v-2 = 24}$

$\implies \rm{v = 24 + 2}$

$\orange{\boxed{\implies \rm{v = 26}}}$

$\large {\sf {Therefore}}$

$\pink{\sf {Velocity\; at \;the\; end\; of\; 2s\; is\; 26\,m/s}}$