Question pattern: Motion graph construction & translation
Original question 1 of 2 in this pattern
From motion description to velocity–time sketch
An object starts from rest. It accelerates uniformly to +6.0 m s⁻¹ during the first 3.0 s, continues at +6.0 m s⁻¹ for 4.0 s, and then changes velocity uniformly to −2.0 m s⁻¹ over the next 4.0 s.
a) Sketch the velocity–time graph from t = 0 s to t = 11 s. Give the coordinates of every vertex and join them with the appropriate straight-line segments.
b) Mark and calculate the time at which the object reverses direction.
Worked answer
Solution
- Step 1
The required vertices are
(0, 0),(3, 6),(7, 6)and(11, -2), with time in seconds and velocity in m s⁻¹. - Step 2
Join the first and last pairs with straight sloping lines and the middle pair with a horizontal line.
- Step 3
The final-segment acceleration is
(-2 - 6) / (11 - 7) = -2.0 m s⁻². - Step 4
Starting from +6.0 m s⁻¹ at t = 7 s, reaching zero takes
6.0 / 2.0 = 3.0 s, so reversal occurs att = 10.0 s.
Final answer
Vertices: (0 s, 0), (3 s, +6.0 m s⁻¹), (7 s, +6.0 m s⁻¹), (11 s, −2.0 m s⁻¹); the graph crosses v = 0 at t = 10.0 s.
Common mistake
Placing the reversal at t = 7 s, when deceleration starts, instead of where velocity becomes zero.
Concepts tested
- velocity–time graph
- uniform acceleration
- constant velocity
- reversal
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