A car moving along a straight highway with speed of $126 \;\mathrm{km} h^{-1}$ is brought to a stop within a distance of $200\; \mathrm{m}$. What is the retardation of the car (assumed uniform in $m/s^2$) ?
A car moving along a straight highway with speed of $126 \;\mathrm{km} h^{-1}$ is brought to a stop within a distance of $200\; \mathrm{m}$. What is the retardation of the car (assumed uniform in $m/s^2$) ?
$u=126 \mathrm{km} / \mathrm{h}=126 \times \frac{5}{18} \mathrm{m} / \mathrm{s}=35 \mathrm{m} / \mathrm{s}$
$v=0$
$s=200 m$
Newton's Equation of motion $v^{2}-u^{2}=2 a s$
$0^{2}-35^{2}=2 a(200)$
$a=-3.0625 \mathrm{m} / \mathrm{s}^{2}$
Similar Questions
The velocity $v$ of a particle as a function of its position $(x)$ is expressed as $v=\sqrt{c_1-c_2 x}$, where $c_1$ and $c_2$ are positive constants. The acceleration of the particle is
If a car at rest accelerates uniformly to a speed of $144 \,km/h$ in $20 \,s$. Then it covers a distance of........$m$
If a car at rest accelerates uniformly to a speed of $144 \,km/h$ in $20 \,s$. Then it covers a distance of........$m$
- [AIPMT 1997]
A car moving with a velocity of $10 \,m/s$ can be stopped by the application of a constant force $F$ in a distance of $20\, m$. If the velocity of the car is $30\, m/s$, it can be stopped by this force in......$m$
A particle starting from rest and moves with constant acceleration travels a distance $x$ in first $2$ seconds and a distance $y$ in next two seconds, then
A particle starting from rest and moves with constant acceleration travels a distance $x$ in first $2$ seconds and a distance $y$ in next two seconds, then
A car, moving with a speed of $50 \,km/hr$, can be stopped by brakes after at least $6\,m$. If the same car is moving at a speed of $100 \,km/hr$, the minimum stopping distance is..........$m$
- [AIEEE 2003]