Summary and Practice Applications of Physics Skills
© by William Dietsch 2000
• Kinematics is the study of motion. No regard is given to the nature of the moving object! All moving things are dealt with by kinematics equally. The cause of the motion is also not important in the study of kinematics.
• Motion is evident when a body changes its position relative to some fixed reference point.
• Displacement is the change of the position of a body in a particular direction. Displacement is a vector quantity, that is it has both magnitude and direction.
• Distance is the “ground covered” by a moving object. Distance only has magnitude (direction is disregarded) and therefore is a scalar quantity.
• The time rate of motion is called speed. Speed is computed by distance/time. Speed is a scalar quantity because any direction is disregarded. The primary unit for speed in the System International is meters per second (m/s). When motion has a direction associated with its speed, it becomes the vector quantity called velocity.
• The term acceleration is used for the rate of change of velocity. When a body speeds up it undergoes + acceleration, if it slows down it has – acceleration. Changing the direction of motion is also acceleration. Changing the speed or direction of a moving body causes acceleration.
Mathematical analysis of motion:
• The speed or velocity of a body is found using v= d/t, where v is speed or velocity (preferably measured in m/s), d is distance (often measured in m), and t is the elapsed time (typically measured in s).
• The acceleration of a body is found using: a = ( v2 - v1 ) / t, where a is the acceleration (measured in m/s2), v1 is the initial velocity (measured in m/s), v2 is the final velocity (must be in the same units), and t is the elapsed time (typically measured in s). If the body starts at rest, the initial velocity, v1, is zero. If the body comes to rest, the final velocity, v2, is zero. If the algebraic sign for the value of the acceleration is +, the body speeds up. If the algebraic sign for the value of the acceleration is -, the body slows down (presuming the body's motion is in the direction defined as positive).
Useful kinematics equations
|a = (v2 - v1) / t or v2 = v1 + at
||v1 v2 a t
|d = ½ (v1 + v2) t
||v1 d v2 t
|d = ½ at2 + v1t
|| v1 d a t
v22 = v12 + 2ad
||v1 d v2 a
Choose an appropriate equation above to solve each of the practice problems below.
Hints for the solution of kinematics problems:
- Read the problem and identify the given and asked data. List the variables in the space provided for the problem solution. Make sure that the units are correct and consistent with each other.
- Select the equation, which includes all of the listed variables from step one.
- Solve the equation for the asked variable.
- Substitute numerical values for the variables and solve.
- Include the proper units, check for significant figures, and highlight (typically circle) the answer.
Graphical representation and analysis of motion:
Graphs are often used to represent a record of the motion of an object. The graphs may subsequently be analyzed to determine speed, velocity, distance, and acceleration. The equations shown above were derived from the graphic analysis of motion!
Distance vs time graphs
• By convention, distance is plotted on the (vertical) Y-axis and elapsed time is plotted on the (horizontal) X-axis, creating a visual record of motion.
• When the d verses t graph yields a straight line, constant velocity is indicated. The slope of the line indicates the average speed of the moving object.
• When the d verses t graph yields a parabola, constant acceleration is indicated. The slope taken at any point using a tangent line at that point yields the instantaneous velocity at that point. If the curve's slope increases, the acceleration has a + sign and indicates that the body is speeding up. If the curve's slope decreases, the acceleration has a - sign indicating that the body is slowing down.
Velocity vs time graphs
• Plotting velocity on the Y-axis and time is plotted on the X-axis, creates an alternated visual record of motion.
• The plot of v verses t of a body moving at constant velocity is a straight line parallel to the x-axis. This occurs because the value of v (the ordinate value is now speed) does not change over time.
• The plot of v verses t of a body accelerating at a constant rate is a straight line with a slope indicative of the value of acceleration. Negative slope portrays a body slowing down and a positive slope portrays it speeding faster.
• The area under a portion of the v verses t curve indicates the displacement (distance traversed) of the body as the result of the motion during the time interval selected.
Kinematics practice problems
- A high school athlete runs 1.00 x 102 m in 12.20 s. What is the velocity in m/s and km/h?
- A person walks 13 km in 2.0 h. What is the person's average velocity in km/h and m/s?
- An Indy-500 racecar's velocity increases from +4.0 m/s to +36 m/s over a 4.0-s period. What is its average acceleration?
- The same racecar slows from +36 m/s to +15 m/s over 3.0 s. What is its average acceleration over this time interval?
- A car is coasting backwards down a hill at -3.0 m/s when the driver gets the engine started. After 2.5 s, the car is moving uphill at a velocity of +4.5 m/s. What is the car's average acceleration?
- A bus is moving at 25 m/s. The driver steps on the brakes, and the bus stops in 3.0 s. a) What is the average acceleration of the bus while braking? b) Suppose the bus took twice as long to stop. How would the acceleration compare to the acceleration you found above?
- A golf ball rolls up a hill toward a Putt-Putt hole. a) If it starts with a velocity of +2.0 m/s and accelerates at a constant rate of -0.50 m/s2, what is its velocity after 2.0 s? b) If the acceleration occurs for 6.0 s, what is its final velocity? c) Describe, in words, the motion of the golf ball.
- A bus traveling at +30 km/h accelerates at a constant +3.5 m/s2 for 6.8 s. What is its final velocity in km/h?
- If a car accelerates from rest at a constant 5.5 m/s2, how long will be required to reach 28 m/s?
- A car slows from 22 m/s to 3 m/s with a constant acceleration of -2.1 m/s2. How long does it require?
- A racecar traveling a +44 m/s is uniformly accelerated to a velocity of +22 m/s over an 11-s interval. What is its displacement during this time?
- A rocket traveling a +88 m/s is accelerated uniformly to +132 m/s over a 15-s interval. What is its displacement during this time?
- A car accelerates at a constant rate form 15 m/s to 25 m/s while it travels 125 m. How long does this motion take?
- A bike rider accelerates constantly to a velocity of 7.5 m/s during 4.5 s. The bike's displacement is +19 m. What was the initial velocity of the bike?
- An airplane starts from rest and accelerates at a constant +3.00 m/s2 for 30.0 s before leaving the ground. What is its displacement during this time?
- Starting from rest, a racecar moves 110 m in the first 5.0 s of uniform acceleration. What is the car's acceleration?
- A driver brings a car traveling at +22 m/s to a full stop in 2.0 s. Assume its acceleration is constant. a) What is the car's acceleration? b) How far does it travel before stopping?
- A biker passes a lamppost at the crest of a hill at +4.5 m/s. She accelerates down the hill at a constant rate of +0.40 m/s2 for 12 s. How far does she move down the hill during this time?
- An airplane accelerates from a velocity of 21 m/s at the constant rate of 3.0 m/s2 over +535 m. What is its final velocity?
- The pilot stops the same plane in 484-m using a constant acceleration of -8.0 m/s2. How fast was the plane moving before braking began?
- A person wearing a shoulder harness can survive a car crash if the acceleration is smaller than -300 m/s2. Assuming constant acceleration, how far must the front end of the car collapse if it crashes while going 101 km/h?
- A car is initially sliding backwards down a hill at -25 km/h. The driver guns the car. By the time the car's velocity is +35 km/h, it is +3.2 m from its starting point. Assuming the car was uniformly accelerated, find the acceleration.
- A brick falls freely form a high scaffold. a) What is its velocity after 4.0 s? b) How far does the brick fall during the first 4.0 s?
- Now that you know about acceleration, test your reaction time. Ask a friend to hold a ruler just even with the top of your fingers. Then have your friend drop the ruler. Taking the number of centimeters that the ruler falls before you can catchy it, calculate your reaction time. An average of several trials will give more accurate results. The reaction time for most people is more than 0.15 s.
- If you drop a golf ball, how far does it fall in 0.5 s?
- A spacecraft traveling at a velocity of +1210 m/s is uniformly accelerated at -150 m/s2. If the acceleration lasts for 8.68 s, what is the final velocity of the craft? Explain your results in words.
- A man falls 1.0 m to the floor. a) How long does the fall take? b) How fast is he going when he hits the floor?
- On wet pavement, a car can be accelerated with a maximum acceleration a = 0.20 g before its tires slip. a) Starting from rest, how fast is it moving after 2.0 s? b) How far has it moved after 4.0 s?
- An express train, traveling at 36.0 m/s is accidentally switched to a sidetrack. The engineer spots a local train exactly 100.0 m ahead on the same track, moving in the same direction at a constant speed of 11.0 m/s. The engineer jams on the brakes and slows the train at a rate of -3.00 m/s2. If the local train keeps going at a steady speed, will the express train stop in time? First, find out how long the express train takes to stop. Then compute how far the express train will take to stop (measured from the point of braking). If this distance exceeds the distance that the local train moves (plus 100 m), then a collision will occur.
- If a bullet leaves the muzzle of a rifle at 600.0 m/s, and the barrel of the rifle is .850 m long, what is the rate of acceleration experienced by the bullet while in the barrel?
- A driver of a car moving at 90.0 km/h suddenly sees a barrier 40.0-m ahead. If the reaction time of the driver is .75 s, and the car can brake at -10.0 m/s2, does the car hit the barrier or stop in time? Compute the maximum speed that the car can be moving and still stop in time.
- A car moving with a constant acceleration covers the distance between two points 60m apart in 6.0s. Its velocity as it passes the second point is 15 m/s. What is its speed at the first point? What is the acceleration? How far behind the first point was the car at rest?
- As a traffic light turns green, a waiting car (at rest) starts with an acceleration of 6.0 m/s2. At the instant the car begins to accelerate, a truck with a constant velocity of 21 m/s passes the car in the next lane. How far will the truck travel before the car catches up with it? What is the speed of the car when it reaches the truck?
- Suppose that you are late to catch a bus. You are 10.0 m away from the bus stop when you see it pull away. It accelerates from rest at a rate of .90 m/s2. Assuming that you can run at constant speed, what is the minimum speed, with which you can run and still catch the bus?
- Plot braking distance versus initial velocity.
|Initial velocity (m/s)
||Braking distance (m)
- Plot braking distance versus the square of the initial velocity.
- Compute the slope of the graph from part b. Find the value and units of the reciprocal of the slope just found. Does this agree with the equation v12 = -2ad? What is the value of a?