If the displacement of a particle at a point | vedclass

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

Question

(D) Given the displacement function $s = 3t^{2} - 12t + 14$. Velocity $v$ is the rate of change of displacement with respect to time $t$, given by $v

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Given the displacement function $s = 3t^{2} - 12t + 14$. Velocity $v$ is the rate of change of displacement with respect to time $t$, given by $v = \frac{ds}{dt}$. $v = \frac{d}{dt}(3t^{2} - 12t + 14) = 6t - 12$. When the velocity becomes zero, we set $v = 0$: $6t - 12 = 0 \implies 6t = 12 \implies t = 2 	ext{ seconds}$. Now, substitute $t = 2$ into the displacement equation to find the displacement at that instant: $s = 3(2)^{2} - 12(2) + 14$ $s = 3(4) - 24 + 14$ $s = 12 - 24 + 14 = 2 	ext{ units}$. Thus, the displacement of the particle when its velocity is zero is $2 	ext{ units}$.

If the displacement of a particle at a point is given by $s = 3t^{2} - 12t + 14$, then the displacement of the particle when its velocity becomes zero is (in $\text{ units}$)

Similar Questions

The equation of motion of a car is $s = t^2 - 2t$,where $t$ is measured in hours and $s$ in kilometers. When the distance travelled by the car is $15 \, km$,the velocity of the car is ......... $km/h$.

The sides of an equilateral triangle are increasing at the rate of $4 \text{ cm/sec}$. Find the rate at which its area is increasing when the side is $14 \text{ cm}$.

If the distance $s$ travelled by a particle in time $t$ is given by $s=t^2-2t+5$,then its acceleration is

If the volume of a sphere increases at the rate of $2 \pi \text{ cm}^3/\text{s}$,then the rate of increase of its radius (in $\text{cm}/\text{s}$),when the volume is $288 \pi \text{ cm}^3$,is

$A$ particle moves in a straight line such that $s = \sqrt{t}$,then its acceleration is proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module