
A motor car runs with a speed of $7\,\,m{s^{ - 1}}$ is stopped by applying brakes. Before stopping, the motor car moves through a distance of $10\,\,m$. Show the total resistance of the motion of the car is one-fourth of the weight of the car.
A) $\dfrac{2}{6}$ of its weight.
B) $\dfrac{2}{4}$ of its weight.
C) $\dfrac{1}{5}$ of its weight.
D) $\dfrac{1}{4}$ of its weight.
Answer
136.5k+ views
Hint: The given problem the acceleration can be solved using the formula that is derived from the equation of motion which incorporates the final velocity of the car, initial velocity of the car and the distance travelled by the car when the brakes are applied.
Useful formula:
The formula for second equation of motion;
${v^2} = {u^2} + 2as$
Where, $v$ denotes the final velocity of the car, $u$ denotes the initial velocity of the car, $a$ denotes the acceleration of the car, $s$ denotes the distance travelled by the car.
Complete step by step solution:
The data given in the problem;
Final velocity of the car, $v = 0\,\,m{s^{ - 1}}$
Initial velocity of the car is, $u = 7\,\,m{s^{ - 1}}$.
Distance travelled by the car, $s = 10\,\,m$.
The formula for second equation of motion;
${v^2} = {u^2} + 2as$
$a = \dfrac{{{v^2} - {u^2}}}{{2s}}$
Substitute the values of final velocity, initial velocity and speed of the car on the formula;
$a = \dfrac{{0\,\,m{s^{ - 1}} - {7^2}\,\,m{s^{ - 1}}}}{{2 \times 10\,\,m}}$
$a = \dfrac{{ - 49}}{{20}}$
$a = - 2.45\,\,m{s^{ - 2}}$
Where $a = - 2.45\,\,m{s^{ - 2}}$ can be written as;
$\dfrac{{49}}{{20}} = \dfrac{{98}}{{40}}$
That is;
$\dfrac{{9.8}}{4} = \dfrac{{ - g}}{4}$
The resistance force that is acting on the car is;
Where, $m$ denotes the mass of the object;
That is the resistance force that is acting on the car is one fourth to the car.
Hence, the option (D), $\dfrac{1}{4}$ of its weight is the correct answer.
Note: The acceleration of the car can be related to the formula $F = ma$ that is the formula derived from Newton's second law of motion. Newton's Second Law of Motion is expressed in such a way that acceleration takes place when a force acts on an object.
Useful formula:
The formula for second equation of motion;
${v^2} = {u^2} + 2as$
Where, $v$ denotes the final velocity of the car, $u$ denotes the initial velocity of the car, $a$ denotes the acceleration of the car, $s$ denotes the distance travelled by the car.
Complete step by step solution:
The data given in the problem;
Final velocity of the car, $v = 0\,\,m{s^{ - 1}}$
Initial velocity of the car is, $u = 7\,\,m{s^{ - 1}}$.
Distance travelled by the car, $s = 10\,\,m$.
The formula for second equation of motion;
${v^2} = {u^2} + 2as$
$a = \dfrac{{{v^2} - {u^2}}}{{2s}}$
Substitute the values of final velocity, initial velocity and speed of the car on the formula;
$a = \dfrac{{0\,\,m{s^{ - 1}} - {7^2}\,\,m{s^{ - 1}}}}{{2 \times 10\,\,m}}$
$a = \dfrac{{ - 49}}{{20}}$
$a = - 2.45\,\,m{s^{ - 2}}$
Where $a = - 2.45\,\,m{s^{ - 2}}$ can be written as;
$\dfrac{{49}}{{20}} = \dfrac{{98}}{{40}}$
That is;
$\dfrac{{9.8}}{4} = \dfrac{{ - g}}{4}$
The resistance force that is acting on the car is;
Where, $m$ denotes the mass of the object;
That is the resistance force that is acting on the car is one fourth to the car.
Hence, the option (D), $\dfrac{1}{4}$ of its weight is the correct answer.
Note: The acceleration of the car can be related to the formula $F = ma$ that is the formula derived from Newton's second law of motion. Newton's Second Law of Motion is expressed in such a way that acceleration takes place when a force acts on an object.
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