The position x of a particle with respect to | vedclass

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

Question

(A) Given the position equation: $x = 9t^2 - t^3$. Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(A) Given the position equation: $x = 9t^2 - t^3$. Velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(9t^2 - t^3) = 18t - 3t^2$. To find the time at which speed is maximum,we differentiate velocity with respect to time and set it to zero: $\frac{dv}{dt} = 18 - 6t = 0$. Solving for $t$,we get $t = 3 \ s$. Now,substitute $t = 3 \ s$ into the position equation to find the position at that instant: $x = 9(3)^2 - (3)^3 = 9(9) - 27 = 81 - 27 = 54 \ m$.

The position $x$ of a particle with respect to time $t$ along the $x$-axis is given by $x = 9t^2 - t^3$,where $x$ is in metres and $t$ is in seconds. What will be the position of this particle when it achieves maximum speed along the $x$-direction?

Similar Questions

The figure shows a velocity-time graph of a particle moving along a straight line. The correct acceleration-time graph of the particle is shown as:

Which one of the following graphs for a body moving along a straight line is possible?

$A$ particle shows a distance-time curve as shown in the figure. The maximum instantaneous velocity of the particle is around the point:

The $x-t$ equation is given as $x = 2t + 1$. The corresponding $v-t$ graph is:

$A$ body starts from rest and travels a distance $S$ with uniform acceleration,then moves uniformly a distance $2S$ and finally comes to rest after moving further $3S$ under uniform retardation. The ratio of the average velocity to maximum velocity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module