The equation of motion of a particle moving a | vedclass

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

Question

(A) Given the equation of motion: $s = 2t^3 - 9t^2 + 12t$. To find the velocity $v$,we differentiate $s$ with respect to $t$: $v = \frac{ds}{dt} = \fr

Vedclass · Accepted Answer

$3/2 	ext{ } sec$

Vedclass · Answer

(A) Given the equation of motion: $s = 2t^3 - 9t^2 + 12t$. To find the velocity $v$,we differentiate $s$ with respect to $t$: $v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 9t^2 + 12t) = 6t^2 - 18t + 12$. To find the acceleration $a$,we differentiate the velocity $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 18t + 12) = 12t - 18$. For the acceleration to be zero,we set $a = 0$: $12t - 18 = 0$. Solving for $t$: $12t = 18$,which gives $t = \frac{18}{12} = \frac{3}{2} 	ext{ } sec$. Thus,the acceleration of the particle will be zero after $3/2 	ext{ } sec$.

The equation of motion of a particle moving along a straight line is $s = 2t^3 - 9t^2 + 12t$,where the units of $s$ and $t$ are $cm$ and $sec$. The acceleration of the particle will be zero after:

Similar Questions

If a particle moves in a straight line according to the law $x = a \sin (\sqrt{\lambda} t + b)$,then the particle will come to rest at two points whose distance is [symbols have their usual meaning]

The edge of a cube is increasing at the rate of $5 \, cm/\sec$. How fast is the volume of the cube increasing when the edge is $12 \, cm$ long? (in $cm^3/\sec$)

The surface area of a spherical balloon being inflated increases at a constant rate. If initially,the radius of the balloon is $3$ units and after $5$ seconds,it becomes $7$ units,then its radius after $9$ seconds is

The rate of change of the volume of a sphere with respect to its surface area when the radius is $5 \ m$ is:

Find the rate of change of $\sqrt{x^2 + 16}$ with respect to $\frac{x}{x - 1}$ at $x = 3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module