Physics: Kinematics: Velocity-Time and Position-Time Graphs Worksheet

Question 1

A position-time graph shows a straight line from position 0 m at time 0 s to position 20 m at time 4 s. Find the object's velocity.

Accepted Answer

The object's velocity is 5 m/s. The slope of a position-time graph is velocity, so velocity = change in position divided by change in time = 20 m / 4 s = 5 m/s.

Question 2

A position-time graph is a horizontal line at 12 m from 0 s to 5 s. Describe the object's motion and find its velocity.

Accepted Answer

The object is not moving because its position stays constant at 12 m. Its velocity is 0 m/s.

Question 3

A position-time graph has points (0 s, 0 m), (2 s, 6 m), (4 s, 12 m), and (6 s, 18 m). Describe the motion and calculate the velocity.

Accepted Answer

The object moves in the positive direction at a constant velocity. The velocity is 3 m/s because the position increases by 6 m every 2 s.

Question 4

A velocity-time graph shows a constant velocity of 3 m/s from 0 s to 8 s. Find the displacement during this time interval.

Accepted Answer

The displacement is 24 m. On a velocity-time graph, displacement equals the area under the graph, so displacement = 3 m/s x 8 s = 24 m.

Question 5

A velocity-time graph starts at 0 m/s at 0 s and increases in a straight line to 20 m/s at 5 s. Find the acceleration and the displacement from 0 s to 5 s.

Accepted Answer

The acceleration is 4 m/s^2 because acceleration = change in velocity divided by time = 20 m/s / 5 s. The displacement is 50 m because the area under the graph is a triangle with area = 1/2 x 5 s x 20 m/s = 50 m.

Question 6

A velocity-time graph shows +6 m/s from 0 s to 4 s, then -2 m/s from 4 s to 7 s. Find the total displacement and the total distance traveled.

Accepted Answer

The total displacement is 18 m. The object moves +24 m in the first 4 s and -6 m in the next 3 s, so displacement = 24 m - 6 m = 18 m. The total distance traveled is 30 m because distance counts both parts as positive: 24 m + 6 m = 30 m.

Question 7

A position-time graph decreases in a straight line from 30 m at 0 s to 0 m at 6 s. Find the velocity and explain the sign of your answer.

Accepted Answer

The velocity is -5 m/s. The slope is (0 m - 30 m) / (6 s - 0 s) = -30 m / 6 s = -5 m/s. The negative sign means the object is moving in the negative direction.

Question 8

A curved position-time graph gets steeper as time increases, and the curve bends upward. Describe what is happening to the object's velocity.

Accepted Answer

The object's velocity is increasing in the positive direction. On a position-time graph, a steeper slope means a greater velocity, and an upward-curving graph shows the slope increasing over time.

Question 9

Runner A's position changes from 0 m to 40 m in 10 s. Runner B's position changes from 10 m to 50 m in 10 s. Compare their velocities.

Accepted Answer

Both runners have the same velocity of 4 m/s. Runner A's velocity is 40 m / 10 s = 4 m/s, and Runner B's velocity is (50 m - 10 m) / 10 s = 4 m/s.

Question 10

A velocity-time graph is a straight line from -4 m/s at 0 s to +4 m/s at 4 s. Find the acceleration and the total displacement from 0 s to 4 s.

Accepted Answer

The acceleration is 2 m/s^2 because the velocity changes by 8 m/s in 4 s. The total displacement is 0 m because the negative area from 0 s to 2 s cancels the positive area from 2 s to 4 s.

Question 11

A car's velocity-time graph has three parts: it speeds up from 0 m/s to 12 m/s from 0 s to 3 s, moves at 12 m/s from 3 s to 7 s, and slows to 0 m/s from 7 s to 9 s. Find the total displacement.

Accepted Answer

The total displacement is 78 m. The first area is a triangle: 1/2 x 3 s x 12 m/s = 18 m. The middle area is a rectangle: 4 s x 12 m/s = 48 m. The last area is a triangle: 1/2 x 2 s x 12 m/s = 12 m. The total is 18 m + 48 m + 12 m = 78 m.

Question 12

An object's position-time graph rises from 0 m to 15 m, becomes flat for a short time, and then slopes downward back toward 0 m. Describe the motion in words.

Accepted Answer

The object first moves in the positive direction, then stops for a short time, and then moves in the negative direction back toward the starting position.

Question 13

For each graph type, state what the slope or area represents: slope of a position-time graph, slope of a velocity-time graph, and area under a velocity-time graph.

Accepted Answer

The slope of a position-time graph represents velocity. The slope of a velocity-time graph represents acceleration. The area under a velocity-time graph represents displacement.

Question 14

A position-time table shows these values: at 0 s the position is 5 m, at 2 s the position is 5 m, at 4 s the position is 9 m, and at 6 s the position is 13 m. During which time interval is the object stopped?

Accepted Answer

The object is stopped from 0 s to 2 s because its position stays at 5 m during that interval. A constant position means the velocity is 0 m/s.

Question 15

An object starts at position 0 m. Its velocity is +2 m/s from 0 s to 3 s, 0 m/s from 3 s to 5 s, and -1 m/s from 5 s to 9 s. Describe the shape of the position-time graph and give the final position.

Accepted Answer

The position-time graph first rises in a straight line from 0 m to 6 m, then stays flat at 6 m from 3 s to 5 s, and then slopes downward to 2 m by 9 s. The final position is 2 m.

Physics: Kinematics: Velocity-Time and Position-Time Graphs

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