Question

A car is moving along a straight highway with a speed of $126 km/hr$ is brought to a stop within a distance of $200m$. What is the retardation of the car (assuming uniform), and how long does it take for the car to stop?

Answer 1

To find the retardation and the time taken, we can use newton’s laws of the equation. First, we can find the acceleration, using the given data for distance, initial and final velocity from $v^{2}-u^{2}=2as$. Using acceleration, we can calculate the time taken from $v=u+at$.

Formula used:
$v=u+at$
$v^{2}-u^{2}=2as$

Complete step by step answer:
Newton’s laws of motion, describe the relationship between the force on a body and its motion. These are the foundations of classical mechanics, also known as Newtonian mechanics. There are three laws:
1. An object remains in its state of motion or rest until an external force acts on it.
2. The force applied on an object is equal to the product of its mass and acceleration.
3. Every action has an equal and opposite reaction.
Since the car is moving in a straight line, we can use Newton’s laws to find the retardation or negative acceleration and the time is taken.
Given that initial velocity $u=126km/hr=126\times \dfrac{5}{18}m/s=35m/s$, final velocity $v=0$ and distance covered $s=200m$.
Then from newton’s laws of the equation, $v^{2}-u^{2}=2as$
Substituting, we get, $0^{2}-35^{2}=2a(200)$
Then $a=-3.06m/s^{2}$
Here the –ve sign indicates that the car is retarding.
Also, we know that, $v=u+at$ again from newton’s law of equation,
Then, substituting we get, $0=35+(-3.06t)$
$t=11.4s$
Thus, the retardation is $3.06m/s^{2}$ and time take is $11.4s$

Note:
Newton’s laws of motion, describe the relationship between the force on a body and its motion. There are three laws. Here since the car is moving in a straight line, we can use Newton’s laws to find the retardation which is negative acceleration i.e. here, $a=-3.06m/s^{2}$ the –ve sign indicates that the car is retarding. It is better to convert speed from $km/hr$ to $m/s$, by $km/hr=\dfrac{5}{18}m/s$