Question

if a=t^2+t+1 find v at t=2s , if at t=1s body is moving with 2m/s in negative -x direction​

Answer 1

Answer:

The velocity of the body at t=2s is 2.83m/s.

Explanation:

Given that,

The acceleration of body as an parameter of time is $a=t^{2}+t+1$

We know the acceleration of a body is the rate of change of its velocity.Therefore,we can write

$a=\frac{dv}{dt} = > dv=adt$

Integrating it on both sides,we get

$\int dv=\int adt= > v=\int (t^{2}+t+1)dt\\= > v=\frac{t^{3}}{3}+ \frac{t^{2}}{2} +t+c$

Also given that,

The velocity of the body at t=1 sec is 2m/s in negative x-direction

$= > v_{t=1}=-2m/s\\= > \frac{1^{3}}{3}+ \frac{1^{2}}{2} +1+c=-2\\= > c=\frac{-23}{6}$

Hence the equation for velocity becomes

$v=\frac{t^{3}}{3}+ \frac{t^{2}}{2} +t-\frac{23}{6}$

Now the velocity of the body at t=2s is

$v_{t=2}=\frac{2^{3}}{3}+ \frac{2^{2}}{2} +2-\frac{23}{6} =\frac{17}{6} =2.83m/s$

Therefore,the velocity of the body at t=2s is 2.83m/s.

#SPJ2

Answer 2

Answer:

The value of $v$ is $\frac{t^3}{3}+\frac{t^2}{2}+t -\frac{26}{3}$.

Explanation:

Given:- The acceleration, $a=t^2+t+1$ and at t = 1s, body is moving with 2 m/s in the negative x-direction.

To find:-

The velocity $v$ at t = 2s.

Step 1 of 2

As we know,

The acceleration is defined as the change in the velocity with respect to the time,

⇒ $a=\frac{dv}{dt}$

⇒ $dv = adt$

Integrate both the sides as follows:

⇒ $\int\limits {dv}\ = \int\limits {adt}$ . . . . . (1)

Now,

Substitute the value $t^2+t+1$ for $a$ in the integral (1) as follows:

⇒ $\int\limits {dv}\ = \int\limits {(t^2+t+1)dt}$

⇒ $v=\frac{t^3}{3}+\frac{t^2}{2}+t +C$ . . . . . (2)

where $C$ is the integration constant.

Step 2 of 2

Since the body is moving with 2 m/s in the negative x-direction at t = 1s.

So, $v=-2m/s$, t = 2s

Substitute the values of $v$ and t in the equation (2) as follows:

⇒ $-2=\frac{2^3}{3}+\frac{2^2}{2}+2 +C$

⇒ $-2=\frac{8}{3}+2+2 +C$

⇒ $C=-2-4-\frac{8}{3}$

⇒ $C=-\frac{26}{3}$

Form (2), we get,

$v=\frac{t^3}{3}+\frac{t^2}{2}+t -\frac{26}{3}$

Final answer: The value of $v$ is $\frac{t^3}{3}+\frac{t^2}{2}+t -\frac{26}{3}$.

#SPJ2