$A$ particle moves along a straight line according to the law $s=16-2t+3t^{3}$,where $s$ metres is the distance of the particle from a fixed point at the end of $t$ seconds. The acceleration of the particle at the end of $2 \ s$ is
- A$3.6 \ m/s^{2}$
- B$36 \ m/s^{2}$
- C$36 \ km/s^{2}$
- D$360 \ m/s^{2}$
Explore More
Similar Questions
If the volume of a sphere is increasing at the rate of $12 \text{ cm}^3/\text{sec}$,then the rate (in $\text{cm}^2/\text{sec}$) at which its surface area is increasing,when the diameter of the sphere is $12 \text{ cm}$,is:
The edge of a cube is decreasing at the rate of $0.04 \ cm/sec$. If the edge of the cube is $10 \ cm$,then the rate of decrease of the surface area of the cube is...
What will be the rate of change of the volume of a sphere with radius $r$,with respect to its diameter?
The rate of change of the diagonal $R$ of a square with respect to its area $A$ is:
Medium
View Solution$A$ cylindrical tank of radius $10 \ m$ is being filled with wheat at the rate of $314 \ m^3/h$. The rate at which the depth of the wheat is increasing is: (in $m/h$)
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo