$A$ particle is moving on a straight line,where its position $s$ (in metre) is a function of time $t$ (in seconds) given by $s = at^2 + bt + 6, t \ge 0$. If it is known that the particle comes to rest after $4 \, s$ at a distance of $16 \, m$ from the starting position $(t = 0)$,then the retardation in its motion is

  • A
    $1 \, m/s^2$
  • B
    $\frac{5}{4} \, m/s^2$
  • C
    $\frac{1}{2} \, m/s^2$
  • D
    $\frac{5}{8} \, m/s^2$

Explore More

Similar Questions

Air is discharging from a large spherical balloon at the rate of $4 \,m^3 / min$. The rate at which the surface area is shrinking when the radius of the balloon is $8 \,m$, is

If $x$ and $y$ are sides of two squares such that $y = x - x^2$,then the rate of change of area of the second square with respect to that of the first square is

The length $x$ of a rectangle is decreasing at the rate of $5 \text{ cm/min}$ and the width $y$ is increasing at the rate of $4 \text{ cm/min}$. When $x = 8 \text{ cm}$ and $y = 6 \text{ cm}$,find the rate of change of the area of the rectangle.

Difficult
View Solution

$A$ vessel in the shape of an inverted cone of height $10 \ ft$ and semi-vertical angle $30^{\circ}$ is full of water. Due to a hole at the vertex,the slant height of the water in the vessel is decreasing at a constant rate of $\frac{1}{\sqrt{3}} \ ft/min$. The rate (in $cu. \ ft/min$) at which the volume of water in the vessel is decreasing,when the volume of water is $\frac{8 \pi}{\sqrt{3}} \ cu. \ ft$,is

The displacement of a particle at time $t$ is $s = t^{3} - 4t^{2} - 5t$. The velocity of the particle at $t = 2 \text{ sec}$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo