# Kinematics Practice Problems

On this page, several problems related to kinematics are given. The solutions to the problems are initially hidden, and can be shown in gray boxes or hidden again by clicking “Show/hide solution.” It is advised that students attempt to solve each problem before viewing the answer, then use the solution to determine if their answer is correct and, if not, why. **Remember to include units on all final answers.**

Content that appears in a box similar to this is content that applies only to the AP C curriculum. AP B and AICE students can skip any content contained in these boxes.

### Displacement

- A rock is thrown straight upward off the edge of a balcony that is 5 m above the ground. The rock rises 10 m, then falls all the way down to the ground below the balcony. What is the rock’s displacement?

a.) What is the magnitude of the child’s displacement?

b.) What is the direction of the child’s displacement?

As you can see in this diagram, the displacement is equivalent to the hypotenuse of a right triangle whose legs are 6 m and 3 m long. So, we can calculate the magnitude of the displacement using the Pythagorean theorem:

### Speed and Velocity

- If the child from problem 2 completes his journey in 20 seconds, what is the magnitude of his average velocity?

b.) Is it possible to move with constant velocity but not constant speed?

### Acceleration

- A car drives in a straight line at a constant speed of 60 miles per hour for 5 seconds. Find its acceleration.

### Uniform Acceleration w/ the Big Five

- A particle moves along the x-axis with an initial velocity of 4 m/s and constant acceleration. After 3 seconds, its velocity is 14 m/s. How far did it travel during this interval

*a* = (*v* 2 )/(2* x*) = (40 2 )/(2 * 80) = 1600/160 = 10 m/s 2 , toward the ground.

### Free Fall

- A rock is dropped 80 meters from a cliff. How long does it take to reach the ground?

Since we are not given nor asked for *v*, we can use the Big 5 number 3, adapted to the y direction:

20 2 /(-2 * -10) = 20 2 /20 = 20 m

-10 * 2 – 20 = -20 – 20 = -40 m/s

-0.5 * -10 * 3 2 – 8 * 3 = 5 * 9 – 24 = 45 – 24 = 21 m

### Projectile Motion

- A tennis ball is thrown horizontally with an initial speed of 10 m/s. If it hits the ground after 4 seconds, how far did it drop before hitting the ground?

0.5 * -10 * 4 2 = -5 * 16 = -80 m

0.5 * -10 * 2 2 + 20 * 2 = -5 * 4 + 40 = -20 + 40 = 20 m

a.) For how long will the cannonball be in the air?

b.) How far will it travel horizontally?

60cos(30 o )6 = 312 m

### Kinematics w/ Graphs

- An object’s position during a 10 second time interval is shown by the graph below:

a.) Determine the object’s total distance traveled and displacement.

b.) What is the object’s velocity at the following times: *t* = 1, *t* = 3, and *t* = 6.

c.) Determine the object’s average velocity and average speed from *t* = 0 to *t* = 10.

d.) What is the object’s acceleration at *t* = 5?

The displacement of the object is simply the final position minus the initial position, or -2 – 0 = -2 m.

a.) Determine the object’s total distance traveled and displacement.

b.) At *t* = 0, the object’s position is *x* = 2 m. Find the object’s position at *t* = 2, *t* = 4, *t* = 7, and *t* = 10.

c.) What is the object’s acceleration at the following times: *t* = 1, *t* = 3, and *t* = 6.

d.) Sketch the corresponding acceleration vs. time graph from *t* = 0 to *t* = 10.

The total distance traveled by the object is simply the sum of all these areas: 3 + 6 + 4.5 + 2 + 2 = 17.5 m

The displacement is found in a similar fashion, except areas below the x-axis are considered negative: 3 + 6 + 4.5 – 2 – 2 = 9.5 m

*x*(10) = 2 + 3 + 6 + 4.5 – 2 – 2 = 11.5 m

a.) At which of the marked points is the object’s velocity the greatest? The least?

b.) Is the object’s acceleration positive or negative between points A and B?

d.) Using the function from part c, determine the object’s maximum and minimum positions and velocities within the interval from *t* = 1 to *t* = 6.

*v*(1) = 3 * 1 2 – 19 * 1 + 23 = 3 – 19 + 23 = 7 m/s

*v*(3) = 3 * 3 2 – 19 * 3 + 23 = 27 – 57 + 23 = -7 m/s

*v*(5) = 3 * 5 2 – 19 * 5 + 23 = 75 – 95 + 23 = 3 m/s

*a*(3) = 6 * 3 – 19 = 18 – 19 = -1 m/s 2

*a*(5) = 6 * 5 – 19 = 30 – 19 = 11 m/s 2

*t* = 1.63008 s or *t* = 4.70326 s

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