An automobile is moving on a straight horizontal road with speed u. What is the shortest distance in which the automobile can be stopped? Take the coefficient of static friction between the tyres and the road as \[{{\mu }_{s}}\]
Answer
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Hint: initial speed of the automobile is given and the final speed must be zero because it came to halt after some time.
The coefficient of static friction is given, that is frictional force will come into play.
Complete step by step answer:
Let us use Newton’s equation of motion to solve this problem.
Initial speed=u
Final speed=0
Let the distance covered be s and the acceleration produced be a.
$
{{v}^{2}}-{{u}^{2}}=2as \\
\implies 0-{{u}^{2}}=-2as \\
\implies s=-\dfrac{{{u}^{2}}}{2a} \\
$
Negative sign is for the acceleration because the body is decreasing the velocity.
From Newton’s second law, F= ma
But here this force is frictional force which is given by \[\mu mg\]
Thus, \[\mu mg=ma\]
\[\Rightarrow \mu g=a\]
Using the value of this acceleration in distance found above we get,
\[s=-\dfrac{{{u}^{2}}}{2\mu g}\].
Additional Information:
Here initially the car is moving and finally, it comes to rest. So negative acceleration called retardation takes place. If a body is accelerating then its acceleration is positive and if the body is coming to stop after some time then it is decelerating and its acceleration is negative.
Note:
While doing such kind of problems we have to keep in mind the number of variables which needs to be introduced and what are the given quantities. Also, we have to specifically keep in our mind whether we have to use equations of constant motion or it involves the use of acceleration.
The coefficient of static friction is given, that is frictional force will come into play.
Complete step by step answer:
Let us use Newton’s equation of motion to solve this problem.
Initial speed=u
Final speed=0
Let the distance covered be s and the acceleration produced be a.
$
{{v}^{2}}-{{u}^{2}}=2as \\
\implies 0-{{u}^{2}}=-2as \\
\implies s=-\dfrac{{{u}^{2}}}{2a} \\
$
Negative sign is for the acceleration because the body is decreasing the velocity.
From Newton’s second law, F= ma
But here this force is frictional force which is given by \[\mu mg\]
Thus, \[\mu mg=ma\]
\[\Rightarrow \mu g=a\]
Using the value of this acceleration in distance found above we get,
\[s=-\dfrac{{{u}^{2}}}{2\mu g}\].
Additional Information:
Here initially the car is moving and finally, it comes to rest. So negative acceleration called retardation takes place. If a body is accelerating then its acceleration is positive and if the body is coming to stop after some time then it is decelerating and its acceleration is negative.
Note:
While doing such kind of problems we have to keep in mind the number of variables which needs to be introduced and what are the given quantities. Also, we have to specifically keep in our mind whether we have to use equations of constant motion or it involves the use of acceleration.
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