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Skill Sheet 4.1
Acceleration Problems
This skill sheet will allow you to practice solving acceleration problems. Remember that acceleration is the rate of change in the speed of an object. In other words, at what rate does an object speed up or slow down? A positive value for acceleration refers to the rate of speeding up, and negative value for acceleration refers to the rate of slowing down. The rate of slowing down is also called deceleration. To determine the rate of acceleration, you use the formula: Final speed – Beginning speed Acceleration = ------------------------------------------------------------------------Change in Time
1. Solving acceleration problems Solve the following problems using the equation for acceleration. Remember the units for acceleration are meters per second per second or m/sec2. The first problem is done for you. 1.
A biker goes from a speed of 0.0 m/sec to a final speed of 25.0 m/sec in 10 seconds. What is the acceleration of the bicycle? 25.0 m 0.0 m 25.0 m ---------------- – ---------------------------sec sec sec = 2.5 macceleration = ------------------------------------ = --------------------------2 10 sec 10 sec sec
2.
A skater increases her velocity from 2.0 m/sec to 10.0 m/sec in 3.0 seconds. What is the acceleration of the skater?
3.
While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the automobile’s acceleration? (Remember that a negative value indicates a slowing down or deceleration.)
4.
A parachute on a racing dragster opens and changes the speed of the car from 85 m/sec to 45 m/sec in a period of 4.5 seconds. What is the acceleration of the dragster?
5.
The cheetah, which is the fastest land mammal, can accelerate from 0.0 mi/hr to 70.0 mi/hr in 3.0 seconds. What is the acceleration of the cheetah? Give your answer in units of mph/sec.
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Skill Sheet 4.1 Acceleration Problems
6.
The Lamborghini Diablo sports car can accelerate from 0.0 km/hr to 99.2 km/hr in 4.0 seconds. What is the acceleration of this car? Give your answer in units of kilometers per hour/sec.
7.
Which has greater acceleration, the cheetah or the Lamborghini Diablo? (To figure this out, you must remember that there are 1.6 kilometers in 1 mile.) Be sure to show your calculations.
2. Solving for other variables Now that you have practiced a few acceleration problems, you can rearrange the acceleration formula so that you can solve for other variables such as time and final speed. Final speed = Beginning speed + ( acceleration × time ) Final speed – Beginning speed Time = ------------------------------------------------------------------------Acceleration 1.
A cart rolling down an incline for 5.0 seconds has an acceleration of 4.0 m/sec2. If the cart has a beginning speed of 2.0 m/sec, what is its final speed?
2.
A car accelerates at a rate of 3.0 m/sec2. If its original speed is 8.0 m/sec, how many seconds will it take the car to reach a final speed of 25.0 m/sec?
3.
A car traveling at a speed of 30.0 m/sec encounters an emergency and comes to a complete stop. How much time will it take for the car to stop if its rate of deceleration is -4.0 m/sec2?
4.
If a car can go from 0.0 to 60.0 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50.0 mi/hr?
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Skill Sheet 4.2
Acceleration and Speed-Time Graphs
Acceleration and distance can be calculated from speed-time graphs.
1. Calculating acceleration from a speed-time graph Acceleration is the rate of change in the speed of an object. The graph below shows that object A accelerated from rest to 10 miles per hour in two hours. The graph also shows that object B took four hours to accelerate from rest to the same speed. Therefore, object A accelerated twice as fast as object B. Clearly, the steepness of the line is related to acceleration. This angle is the slope of the line and is found by dividing the change in the y-axis value by the change in the x-axis value. ∆y Acceleration = -----∆x In everyday terms, we can say that the speed of object A “increased 10 miles per hour in two hours.” Using the slope formula: ∆y 10 mph – 0 mph 5 mph Acceleration = ------ = --------------------------------------- = --------------∆x 2 hours – 0 hour hour •
Acceleration = ∆y/∆x (the symbol ∆ means “change in”)
•
Acceleration = (10 mph – 0 mph)/(2 hours – 0 hours)
•
Acceleration = 5 mph/hour (read as 5 miles per hour per hour)
Beginning physics students are often thrown by the double per time label attached to all accelerations. It is not so alien a concept if you break it down into its parts: The speed changes . . .
. . . during this amount of time:
5 miles per hour
per hour
Accelerations can be negative. If the line slopes downward, ∆y will be a negative number because a larger value of y will be subtracted from a smaller value of y.
2. Calculating distance from a speed-time graph The area between the line on a speed-time graph and the baseline is equal to the distance that an object travels. This follows from the rate formula: Rate or Speed = Distance --------------------Time R = D ---T Or, rewritten:. RT = D miles/hour × 3 hours = 3 miles
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Skill Sheet 4.2 Acceleration and Speed-Time Graphs
Notice how the labels cancel to produce a new label that fits the result. Here is a speed-time graph of a boat starting from one place and sailing to another: The graph shows that the sailboat accelerated between the second and third hour. We can find the total distance by finding the area between the line and the baseline. The easiest way to do that is to break the area into sections that are easy to solve and then add them together. A + B + C + D = distance •
Use the formula for the area of a rectangle, A = L × W, to find areas A, B, and D.
•
Use the formula for finding the area of a triangle, A = l × w/2, to find area C. A + B + C + D = distance 10 miles + 5 miles + 1 mile + 21 miles = 37 miles
3. Acceleration from speed-time graph practice Calculate acceleration from each of these graphs. 1.
2.
3.
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Skill Sheet 4.2 Acceleration and Speed-Time Graphs
4.
Find acceleration for segment 1 and segment 2.
4. Distance from speed-time graph practice Calculate total distance from each of these graphs. 1.
2.
3.
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Skill Sheet 4.3
Acceleration Due to Gravity
One formal description of gravity is “The acceleration due to the force of gravity.” The relationships among gravity, speed, and time are identical to those among acceleration, speed, and time. This skill sheet will allow you to practice solving acceleration problems that involve objects that are in free fall.
1. Gravity, velocity, distance, and time When solving for velocity, distance, or time with an object accelerated by the force of gravity, we start with an advantage. The acceleration is known to be 9.8 meters/second/second or 9.8 m/sec2. However, three conditions must be met before we can use this acceleration: •
The object must be in free fall.
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The object must have negligible air resistance.
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The object must be close to the surface of the Earth.
In all of the examples and problems, we will assume that these conditions have been met and therefore acceleration due to the force of gravity shall be equal to 9.8 m/sec2 and shall be indicated by g Because the y-axis of a graph is vertical, change in height shall be indicated by y. Remember that speed refers to “how fast” in any direction, but velocity refers to “how fast” in a specific direction. The sign of numbers in these calculations is important. Velocities upward shall be positive, and velocities downward shall be negative.
2. Solving for velocity Here is the equation for solving for velocity: final velocity = initial velocity + ( the acceleration due to the force of gravity × time ) OR v = v 0 + gt Example: How fast will a pebble be traveling 3 seconds after being dropped? v = v 0 + gt 2
v = 0 + ( –9.8 meters/sec × 3 sec ) v = – 29.4 meters/sec (Note that gt is negative because the direction is downward.)
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Skill Sheet 4.3 Acceleration Due to Gravity
3. Problems 1.
A penny dropped into a wishing well reaches the bottom in 1.50 seconds. What was the velocity at impact?
2.
A pitcher threw a baseball straight up at 35.8 meters per second. What was the ball’s velocity after 2.5 seconds? (Note that, although the baseball is still climbing, gravity is accelerating it downward.)
3.
In a bizarre but harmless accident, Superman fell from the top of the Eiffel Tower. How fast was Superman traveling when he hit the ground 7.8 seconds after falling?
4.
A water balloon was dropped from a high window and struck its target 1.1 seconds later. If the balloon left the person’s hand at –5 meters/sec, what was its velocity on impact?
4. Solving for distance Imagine that an object falls for one second. We know that at the end of the second it will be traveling at 9.8 meters/second. However, it began its fall at zero meters/second. Therefore, its average velocity is half of 9.8 meters/second. We can find distance by multiplying this average velocity by time. Here is the equation for solving for distance. Look to find these concepts in the equation: the acceleration due to the force of gravity × time distance = ---------------------------------------------------------------------------------------------------------------------- × time 2 OR 1 2 y = --- gt 2 Example: A pebble dropped from a bridge strikes the water in exactly 4 seconds. How high is the bridge? 1 2 y = --- gt 2 1 y = --- × 9.8 meters/sec × 4 sec × 4 sec 2 1 2 y = --- × 9.8 meters/sec × 4 sec × 4 sec 2 y = 78.4 meters Note that the terms cancel. The answer written with the correct number of significant figures is 78 meters. The bridge is 78 meters high.
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Skill Sheet 4.3 Acceleration Due to Gravity
5. Problems 1.
A stone tumbles into a mineshaft and strikes bottom after falling for 4.2 seconds. How deep is the mineshaft?
2.
A boy threw a small bundle toward his girlfriend on a balcony 10.0 meters above him. The bundle stopped rising in 1.5 seconds. How high did the bundle travel? Was that high enough for her to catch it?
3.
A volleyball serve was in the air for 2.2 seconds before it landed untouched in the far corner of the opponent’s court. What was the maximum height of the serve?
6. Solving for time The equations demonstrated in Sections 2 and 3 can be used to find time of flight from speed or distance, respectively. Remember that an object thrown into the air represents two mirror-image flights, one up and the other down. Original equation Time from velocity Time from distance
Rearranged equation to solve for time v–v t = ------------0g
v = v 0 + gt 1 2 y = --- gt 2
t =
2y -----g
Try these: 1.
At about 55 meters/sec, a falling parachuter (before the parachute opens) no longer accelerates. Air friction opposes acceleration. Although the effect of air friction begins gradually, imagine that the parachuter is free falling until terminal speed (the constant falling speed) is reached. How long would that take?
2.
The climber dropped her compass at the end of her 240-meter climb. How long did it take to strike bottom?
3.
For practice and to check your understanding, use these equations to check your work in Sections 2 and 3.
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