Following are a variety of problems involving uniformly accelerated motion along a line. In the solution a list of known quantities will be given followed by a list of quantities wanted. The equations to be used will be identified by number from the list below, but the algebraic work of solving the equations will be left to the student. In most of the problems there will be a discussion of alternate methods of solution or suggestions and hints as to how to attack this particular type of problem. Do not try to just read this section. You must use pencil, paper and a calculator to obtain the maximum value from studying these problems.
Equation 4: vf = vi + aΔt
Equation 5: Δx = viΔt + a(Δt)2
Equation 6: vf2 = vi2 + 2aΔd
Sample Problem #1
A car starts from rest and accelerates with a uniform acceleration of 10
. Calculate how long it will take for the car to reach a speed of 90
(slightly over 60 mph), and calculate the distance the car moves during this time.
Sample Solution #1
This problem can be solved by a straight forward application of equation 4 to find Δt and equation 5 to find ΔX. The student should demonstrate that the correct answers are Δt=9 s and ΔX=405 ft. It is a good idea to try to find the answers using slightly different procedures whenever possible. For example, in this problem a quick check can be made by using your calculated value for Δt, from equation 4 to find the average speed and then set X=VavgΔt to get 405 ft a different way. A second different method would be to use equation 6 and the value of Δt obtained from equation 4 to calculate ΔX. However, while both of the alternate techniques are good checks on the validity of your calculations it is best to avoid using an answer to one part as a basis for a second calculation whenever possible.
Sample Problem #2
A car moving at 30
stops with a constant acceleration in a distance of 100 m. Calculate the acceleration and the time to stop.
Sample Solution #2
To solve the problem apply equation 5 to find the acceleration (a=-4.5
). Use the calculated value of the acceleration to find the time (Δt=6.67 s) using either equation 4 or 6 (4 is easier).
Sample Problem #3
Explain the significance of the negative sign in problem 2.
Sample Solution #3
The acceleration is in the negative direction. Since V was arbitrarily assigned a positive direction, the acceleration must be in the opposite direction.
Sample Problem #4
Assume that the car in problem 2 keeps the same acceleration for an additional 5 seconds. Find its speed and position at the end of this time.
Sample Solution #4
Notice that here we are not asked for ΔX but are asked for X, the position at the end of the 5th second. We need to be sure to specify the position unambiguously, either with reference to the car's position at the start of problem 2 or with respect to its position at the end of problem 2 and the start of this problem. We will place the origin of the reference system at the position of the car at the start of problem 2. Thus at the end of problem 2 it has a position of X=100 m. Using equation 6 with the data listed above we see that ΔX= -56.25 m. The final position of the car is therefore 100 - 56.25 or 43.75 m. To find the speed use equation 4 and obtain a speed of - 22.5
. The complete answer to the problem could be stated as follows: "The car is 43.75 m from its starting point in the direction it was originally moving. It is moving back toward the starting point at a speed of 22.5
and is accelerating toward the starting point with an acceleration of 4.5
." This statement is much more meaningful than simply writing X = 43.75 m and V = -22.5
Sample Problem #5
A baseball is batted vertically with an initial speed of 45
(a) the time before it returns to the level at which it started.
(b) the maximum height the ball reaches.
(c) the position, speed and acceleration 6 seconds after it is batted.
(d) the position and speed 12 seconds after it is batted.
Sample Solution #5
Read this problem carefully and decide the order that you will do the problem. Keep in mind the facts mentioned in the hints such as: the time to reach the top is equal to one-half the total time in the air; the speed at the top is zero; and the speed when it reaches the same level is the same as the initial speed. In multi part problems such as this one, it will be necessary to list the given quantities for each part. As you solve the problem, check to be sure the answers to the various sections are consistent with each other. Before starting the problem decide which direction to call positive. In this problem we will call the upward direction positive and the downward direction negative.
This is easy. You may use either equation 4 or equation 6. If equation 6 is used, you will obtain two roots, Δt=0 or 9.2 seconds. The first root, 0, corresponds to the first time the ball is at the starting level, when it is hit. The second root is the desired answer, the time the ball returns to the starting level.
0 (speed at top is zero)
Use equation 4 to find Δt = 4.6 s. Note that this is half the total time found in part a, so these two calculations are consistent with each other. Use equation 5 to show that X = 103.3 m. As a check on your work plug the value obtained for Δt into equation 6 and see if the same value is obtained for ΔX, it should be.
No problem here. Use equation 4 to find that the ball has a velocity of -13.8
. The negative sign shows that the ball is moving downward. This is consistent with what we found in part b, that the time to reach the top of the trajectory is 4.6 seconds. At 6 seconds the ball will have reached the top and started moving down so its speed should be negative.Use equation 6 to find ΔX to be 93.6 m. If we take the starting point of the motion to be at X = 0 then the position is equal to ΔX so the ball is 93.6 m above the starting point. Again this is consistent since the displacement is less than the maximum height of 103.3 found in part b. Finally note that equation 6 gives the change in position not the distance moved. The ball has actually traveled further than 93.6 m. It has traveled to the top of its flight, 103.3 m, and back down to a point 93.6 m above the ground, a distance of 9.7 m. Thus, the total distance is 113 m. The problem also asks for the acceleration at t = 6 seconds, but while the ball remains in free fall the acceleration is constant at -9.8
Don't get caught here. You have been asked to find the speed and position at 12 seconds. You might be tempted to repeat the procedure of part c but if you do the answers you get will not be reasonable. The magnitude of the velocity will be greater than 45
and the position will be negative. This can happen only if there is a hole in the ground and the ball has fallen into it and is now below ground level. You should immediately notice that 12 s is beyond the time that the ball takes to return to the original position. The correct answer is that the speed and position of the ball cannot be found from the information given since the ball is no longer in free fall.
Sample Problem #6
A ball dropped from the roof of a tall building passed a window ledge with a speed of 96
and struck the ground 1.0 s later.
(a) What is the height of the window?
(b) How tall is the building?
Sample Solution #6
Note that since all of the motion is downward we have chosen to call the downward direction positive. We have called the speed at the start of the last second, Vi. ΔX can be easily found by applying equation 6.(ans: ΔX = 112 ft) Since this is the distance the ball fell during the last second, it must be the height of the window.
There are several ways to solve this part of the problem. One could calculate the time to fall to the window using equation 4, add 1 second to get the total time, and use the total time to find the distance using equation 6. One could use the information known about the last second of the fall to find the final speed (use equation 5) and use this to find the total distance fallen (use equation 5 again). Another method would be to use the speed at the start of the motion (0
), the speed at the window, and the acceleration of gravity to find the distance from the top of the building to the window. Add 112 ft to get the total height. We must decide how we are going to solve the problem before setting up our table of known quantities. We will do it the last way.
Use equation 5 to find ΔX = 144 ft. Since this is only the height from the top of the building to the window we must add the distance to the ground, 112 ft, to get the height of the building: 256 ft.