Physics: Kinematics: Velocity-Time and Position-Time Graphs
Interpreting slope, area, displacement, and acceleration
Physics: Kinematics: Velocity-Time and Position-Time Graphs
Interpreting slope, area, displacement, and acceleration
Physics - Grade 9-12
- 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.
Use slope = rise / run.
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. - 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.
The object is not moving because its position stays constant at 12 m. Its velocity is 0 m/s. - 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.
Check whether the slope stays the same between each pair of points.
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. - 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.
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. - 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.
Use slope for acceleration and area for displacement.
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. - 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.
Displacement can be positive or negative, but distance is always added as a positive amount.
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. - 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.
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. - 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.
Look at how the slope changes from left to right.
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. - 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.
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. - 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.
Areas below the time axis are negative, and areas above the time axis are positive.
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. - 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.
Break the graph into triangles and rectangles.
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. - 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.
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. - 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.
Match position change with velocity, velocity change with acceleration, and velocity multiplied by time with displacement.
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. - 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?
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. - 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.
Use each velocity to find how much the position changes during that time interval.
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.