$A$ 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 $r$ is the radius of a spherical balloon at time $t$ and the surface area of the balloon changes at a constant rate $K,$ then ....
The volume of a cube is increasing at a rate of $9 \text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is $10 \text{ cm}$ (in $\text{ cm}^2/\text{s}$)?
Medium
View SolutionThe radius of a sphere increases at the rate of $0.04 \text{ cm/sec}$. The rate of increase in the volume of that sphere with respect to its surface area,when its radius is $10 \text{ cm}$ is
The interval in which the function $f(x) = x^3$ increases less rapidly than $g(x) = 6x^2 + 15x + 5$ is:
Difficult
View SolutionThe total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 3x^2 + 36x + 5$. The marginal revenue,when $x = 15$ is . . . . . . .
DifficultGUJCET 2026
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