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Question: Consider a car moving on a straight road with a speed of \[100m/s\] . The distance at which a car ca...

Consider a car moving on a straight road with a speed of 100m/s100m/s . The distance at which a car can be stopped (μk=0.5)\left( {{\mu _k} = 0.5} \right)
A. 100m100m
B. 400m400m
C. 800m800m
D. 1000m1000m

Explanation

Solution

To find the distance at which the car stops, first we must find the acceleration of the car when the brakes are applied. Once we find the acceleration we need to find the distance necessary to stop the car if the acceleration or retardation is split. The expression used is:
v2=u2+2aS{v^2} = {u^2} + 2aS herevv is the final velocity, uu is the initial velocity, aa is the acceleration, and SS is the distance at which it is made still.

Complete step by step solution:
The speed by which the car is moving is known as velocityuu. Now, the car is stationary due to the use of the brakes which makes the final velocity zero. Therefore, the distance to stop is dd and the initial velocity is uu , in turn, makes the acceleration of the car till stop as aa, and to estimate the value of we apply the relation of v2=u2+2aS{v^2} = {u^2} + 2aS. Once we get the acceleration value, we substitute the value of the acceleration in the second case or when the car comes to a stop. In the question when the car is stopped by friction, the initial velocity u=100m/su = 100m/s, the given coefficient μk=0.5{\mu _k} = 0.5 , and g=10m/s2g = 10m/{s^2} .
The retarding force is given as ma=μRma = \mu R
ma=μmgma = \mu mg Which becomes a=μga = \mu g
Let the car cover distance SSwith velocity uu and stop. Hence we know that the relation for finding the stopping distance is given by
v2=u2+2aS{v^2} = {u^2} + 2aS
v21002=2aS\Rightarrow {v^2} - {100^2} = 2aS
u2=2aS\Rightarrow {u^2} = 2aSasv=0v = 0
S=u22a=10022×μg=10022×0.5×10\Rightarrow S = \dfrac{{{u^2}}}{{2a}} = \dfrac{{{{100}^2}}}{{2 \times \mu g}} = \dfrac{{{{100}^2}}}{{2 \times 0.5 \times 10}}
Therefore the stopping distance is S=1000mS = 1000m

Note:
The word retardation means backward or negative acceleration where the acceleration is applied during braking purposes. Thus the stopping distance can be defined as the distance covered by the body before it rests. It is related to various factors like road surfaces, the reflex of the driver, etc. Its SI unit is metermm.