Concept explainers
Section (I)
The acceleration of a car that must be stopped on a dime.
Section (I)
Answer to Problem 15P
Solution:
The acceleration of the car is -2807.4 m/s2
Explanation of Solution
Given:
Initial velocity of the car,
Final velocity of the car,
Diameter of the dime,
Formula:
Third equation of motion is:
Where, v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.
Calculation:
Convert unit of initial velocity from km/h to m/s.
Convert diameter of dime from cm to m.
Substitute the given values in the equation of motion and solve for acceleration (a).
Conclusion: When the car stops on thedime, its acceleration is 2807.4 m/s2 in the backward direction.
Section(II)
The number of g’s in the value of acceleration calculated.
Section(II)
Answer to Problem 15P
Solution:
Acceleration is approximately 286 times the acceleration due to gravity.
Explanation of Solution
Given:
Acceleration of the car,
Formula Used:
Number of g’s =
Where, a is the acceleration and g is the gravitational acceleration.
Calculation:
Substitute the values and solve.
Number of g’s
Conclusion: The acceleration is approximately 286 times the acceleration due to gravity.
Section (III)
The force experienced by an occupant of the car in stopping the car on a dime.
Section (III)
Answer to Problem 15P
Solution:
The force experienced by the occupant is 190903 N
Explanation of Solution
Given:
Mass of the occupant,
Acceleration of the car,
Formula:
From Newton’s second law of motion:
Here, m is the mass and a is the acceleration.
Calculation:
Substitute the values and solve for force ( F ).
Conclusion: Due to this acceleration occupant experiences a force of 190903 N.
Chapter 4 Solutions
Physics: Principles with Applications
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