An object with mass m is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If s(t) is the distance dropped after t seconds, then the speed is v = s'(t) and the acceleration is a = v'(t). If g is the acceleration due to gravity, then the downward force on the object is mg − cv, where c is a positive constant, and Newton's Second Law gives m(dv/dt)=mg-cv Solve this as a linear equation. (Use v for v(t).) What is the limiting velocity? lim t → ∞ Find the distance the

An object with mass m is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If s(t) is the distance dropped after t seconds, then the speed is v = s'(t) and the acceleration is a = v'(t). If g is the acceleration due to gravity, then the downward force on the object is mg − cv, where c is a positive constant, and Newton's Second Law gives m(dv/dt)=mg-cv Solve this as a linear equation. (Use v for v(t).) What is the limiting velocity? lim t → ∞ Find the distance the

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Description: An object with mass m is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If s(t) is the distance dropped after t seconds, then the speed is v = s'(t) and the acceleration is a = v'(t). If g is the a

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter6: Motion In Two Dimensions

Section6.3: Relative Velocity

Problem 27PP

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An object with mass m is dropped from rest and we assume that the air resistance is proportional to the speed of the object. If s(t) is the distance dropped after t seconds, then the speed is v = s'(t) and the acceleration is a = v'(t). If g is the acceleration due to gravity, then the downward force on the object is mg − cv, where c is a positive constant, and Newton's Second Law gives m dv dt = mg − cv. (a) Solve this as a linear equation. (Use v for v(t).) v=mgC(1−e−(Cm)t) (b) What is the limiting velocity? lim t → ∞ v(t) = mgc (c) Find the distance the object has fallen after t seconds. (Use s for s(t).)
When a car is moving at x miles per hour and the driver decides to slam on the brakes, the car will travel x + 1/20 x2 feet. (The general formula is f (x) = ax + bx2, where the constant a depends on the driver’s reaction time and theconstant b depends on the weight of the car and the type of tires.) If a car travels 175 feet after the driver decides to stop, how fast was the car moving? (Source: Applying Mathematics: A Course in Mathematical Modelling.)
A car weighing 2.5 metric tons and traveling at 90 km/h hits a 500 m long stretch of black ice. Unfortunately, due to skidding, neither accelerating nor braking has any effect on the speed! The driver manages to maintain steady straight direction of motion and the only impact is provided by the ice friction force, which is numerically equal to 4v² Newtons, where the velocity v of the car is measured in m/sec. (a) Using Newton's Second Law F = ma, set up a mathematical model for the position x(t) and velocity v(t) of the car as functions of time t. Start by drawing a diagram and choosing a consistent system of units based on kg, m, sec (1 ton = 1000 kg, 1 m/sec = 3.6 km/h, 1 N = 1 kg · m/sec²). Introduce and label the variables, show the units and write down the differential equations and the intial conditions. (b) Use the model in part a to calculate v(t) and x(t). Fully show the process of solving the initial value problems. (c) Based on your work so far, how long will it take to pass…
According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass m that has been projected vertically upward from the earth's surface is mgR? F = (x + R)- where x = x(1) is the object's distance above the surface at time t, R is the earth's radius, and g is the acceler- ation due to gravity. Also, by Newton's Second Law, F = ma = m(dv/dt) and so dv m dt mgR? (x + R)? (a) Suppose a rocket is fired vertically upward with an initial velocity vo. Let h be the maximum height above the surface reached by the object. Show that 2gRh VR+ h [Hint: By the Chain Rule, m (dv/di) = mv (dv/dx).] (b) Calculate v. = lim--- Vo. This limit is called the escape velocity for the earth. (c) Use R = 3960 mi and g = 32 ft/s² to calculate v, in feet per second and in miles per second.
You drop an object of mass m from a tall building. Suppose the only forces affecting its motion are gravity, and air resistance proportional to the object's speed with positive constant of proportionality k. Let g denote gravitational acceleration (a positive constant). Express the total force in terms of m, g, and the object's velocity v, where upward displacement is considered positive. F = mg - kv Newton's second law tells us that force is equal to mass x acceleration, F = ma. Relating acceleration to velocity, rewrite the equation for total force above as a first order differential equation for v as a function of t. Denote v' as dv dt v(t) m(- dr) this is not an equation. Solve this differential equation for v(t) with the initial condition v(0) = V0. = mg (1-e ==) m k Find the terminal Terminal velocity = mg k X velocity. X syntax error: X X
For a mass m falling in air, we assume that the air resistance is proportional to the speed v (t): mv '(t) = mg - kv (t),where g is the acceleration of gravity and k is a positive constant that depends on the geometry of the object.What is v (t) given that v (0) = 0?
Problem 2: A 165 lb skydiver jumps off from a plane at a height 10,000 ft. In the first 30 seconds, the drag force due to air resistance is 2v. Use g = 32 ft/s'. (A) What is the differential equation (as an explicit derivative) that describes the motion of the skydiver? (B) what is the velocity of the skydiver after the first 10 seconds? (C) What is his altitude at that instant of time? After the first 30 seconds, the parachute is released to increase the drag force. If this time, the drag force is proportional to the square of the velocity, (D) find an expression for the velocity at any time t after the initial 30 seconds, assuming the same value of k. It's alright if your answer is in implicit form.
The block has mass m = 6.8 kg, the ramp is at an angle of θ = 28 degrees, the coefficient of kinetic friction between the block and the ramp is μk = 0.25, and the applied force is F = 156.4 N. What is the acceleration of the block up the ramp, in m/s2?
A 2.75 kg ball is dropped straight down on a concrete floor and bounces straight up. At an instant when the ball is in contact with the floor, its acceleration is 34.0 m·s−2 upward. At that instant, calculate the force on the ball that is exerted by the floor.
A man pushes an object to the right and exerts a force which has a horizontal compotent of F = 33 N. A horizontal frictional force has a magnitude of f = 15 N which opposed the horizontal component of the fushing force. The mass of the object is m = 31 kg. Write an expression for the magnitude of the acceleration of the object. If the object starts at rest what is the speed in meters per second at t = 2.00s? If the man stops pushing the object at t = 2.00s and the firctional force is constant what is the distance in meters does to object slide before coming to a rest?
The acceleration of an object moving through a viscous fluid (like honey) is a = -bv / m, where v is the speed of the object and m is its mass. What are the SI units of b? kg /s O kg /s² kg / m kg / m? kg / m·s none of these
A skydiver’s vertical velocity is governed by m(dv/dt) = mg - K(v^2) where, m is the mass of the skydiver, g is gravitational acceleration, and K is the skydiver’s coefficient of drag. (i) If the skydiver leaves his/ her aero plane at time t = 0 with zero vertical velocity, find the vertical velocity equation, v. (ii) Compute the time that the skydiver reaches quarter of his/her final velocity.(iii) Solve the equation of v(t) assuming that you are the skydiver, g = 9.8m/(s^2), and K = 110kg/s. Then, make a plot of the vertical velocity equation, v as a function of time, t for 0 ≤ t ≤ 3.