The velocity of a particle moving along the x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The velocity of the particle is given by $v(t) = 1 - 3t^2 + 2t^3$. We know that velocity is the rate of change of position, $v = \frac{dx}{dt}$. T

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The velocity of the particle is given by $v(t) = 1 - 3t^2 + 2t^3$. We know that velocity is the rate of change of position, $v = \frac{dx}{dt}$. Therefore, the displacement $x$ can be found by integrating the velocity with respect to time: $x(t) = \int v(t) \ dt = \int (1 - 3t^2 + 2t^3) \ dt$. Performing the integration: $x(t) = t - t^3 + \frac{2t^4}{4} + C = t - t^3 + \frac{t^4}{2} + C$. Given that at $t = 0$, $x = 0$, we substitute these values to find the constant $C$: $0 = 0 - 0^3 + \frac{0^4}{2} + C \Rightarrow C = 0$. So, the position function is $x(t) = t - t^3 + \frac{t^4}{2}$. At $t = 2 \ s$, the position is: $x(2) = 2 - (2)^3 + \frac{(2)^4}{2} = 2 - 8 + \frac{16}{2} = 2 - 8 + 8 = 2 \ m$.

The velocity of a particle moving along the $x$-axis varies as a function of time $t$ as $v(t) = (1 - 3t^2 + 2t^3) \ m/s$. If its position at $t = 0$ is $x = 0$, then at $t = 2 \ s$, its position is: (in $m$)

Similar Questions

The stopping distance of a vehicle moving with a given acceleration is proportional to the ....... of its speed.

If the velocity of a particle is $v = At + Bt^2$,where $A$ and $B$ are constants,then the distance travelled by it between $1 \ s$ and $2 \ s$ is

The relation between time $t$ and displacement $x$ is $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,the retardation is:

$A$ body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $5^{th}$ second to that covered in $5$ seconds is

Vedclass Test Series

Exam Paper Generator

Online Exam Module