01 · Falling velocity
Velocity decreases uniformly from10 to2m/s in4s. Find acceleration.
Hint
Use (2−10)/4.
Worked solution
Acceleration=−2m/s². While velocity remains positive, speed is decreasing.
Understand · explore · practise
Velocity–time gradients and acceleration
Calculate signed acceleration from velocity–time gradients and separate direction of motion, speed change and zero velocity.
Before you startGradients, signed velocity and speed.
01 / Read velocity from height and acceleration from slope
A graph below the time axis can still have positive acceleration.
Acceleration is change in velocity per unit time.
A straight velocity-time line has constant acceleration equal to its gradient. A horizontal nonzero line means constant nonzero velocity; only the line v=0 represents rest.
Height is velocity; slope is accelerationExplore
Right is positive. Each straight velocity-time segment runs from t=0 to t=4 seconds. Choose a pair of endpoint velocities, measured in m/s.
02 / Use the change between two points
Subtract both the velocities and the times.
A straight graph passes through (3 s,4 m/s) and (7 s,12 m/s).Worked example
Change in velocity = 12−4 = 8 m/s
Keep the sign of final minus initial.
Elapsed time = 7−3 = 4 s
These are clock readings, not durations from zero.
Acceleration = 8/4 = +2 m/s²
A positive slope is positive acceleration.
02 · Offset times
A straight graph goes from v=−5m/s at t=2s to v=7m/s at t=8s. Find acceleration.
Hint
Subtract coordinates in the same order.
Worked solution
Acceleration=(7−(−5))/(8−2)=12/6=+2m/s².
03 · Gradient unit
Explain why acceleration has units m/s².
Hint
Divide velocity units by time units.
Worked solution
(m/s)/s=m/s². The graph height is measured in m/s, but its slope is measured in m/s².
03 / A zero slope does not mean zero velocity
Check the vertical value separately.
04 · Cruise
A graph is horizontal at v=6m/s for5s. State velocity, speed and acceleration.
Hint
A horizontal line has slope0.
Worked solution
Velocity=+6m/s, speed=6m/s and acceleration=0. The particle moves right at constant speed.
05 · Rest
What is different about a graph horizontal at v=0?
Hint
Now both height and slope are zero.
Worked solution
The particle remains stationary throughout that interval, with zero velocity and zero acceleration.
04 / Compare height and slope to determine speed change
Positive acceleration does not always mean increasing speed.
Watch: equal acceleration, different speed changes
Pause, replay or seek freely. The notes explain the same idea and stay in view.
06 · Negative but rising
Velocity rises uniformly from−8 to−2m/s in4s. Describe the motion and find acceleration.
Hint
The signed value rises while its magnitude falls.
Worked solution
Acceleration=+1.5m/s². The particle moves left throughout, slowing from8 to2m/s.
07 · Negative and falling
Velocity falls uniformly from−2 to−8m/s in4s. Describe it.
Hint
Compare magnitudes.
Worked solution
Acceleration=−1.5m/s². The particle moves left and speeds up from2 to8m/s. Negative acceleration is not automatically slowing down.
05 / Split your description at a zero crossing
Constant acceleration can first slow motion and then speed it up.
08 · Turn time
A straight graph joins (0,6) to (4,−6), with time in seconds and velocity in m/s. When does the particle reverse?
Hint
Find the line or use symmetry.
Worked solution
Acceleration=(−6−6)/4=−3m/s² and v=6−3t. It reaches v=0 at t=2s, then changes from rightward to leftward motion.
09 · At the crossing
Is acceleration zero at t=2s in that model?
Hint
The line retains its slope.
Worked solution
No. Velocity is zero at that instant, while acceleration remains−3m/s².
10 · Whole interval
Describe speed from0 to4s for that model.
Hint
Use |v| on each side of the crossing.
Worked solution
It falls from6 to0m/s during0≤t≤2 and rises from0 to6m/s during2≤t≤4. Acceleration remains constant throughout.
06 / A secant gives average acceleration
A tangent gives instantaneous acceleration on a smooth curve.
11 · Curved graph
For v=t² m/s on0≤t≤2s, calculate average acceleration. Does that mean acceleration is constant?
Hint
Use the endpoint velocity change.
Worked solution
Average acceleration=(4−0)/2=2m/s². It is not constant: the curve gets steeper; its instantaneous acceleration is2t m/s².
12 · Compare graph types
How do you recognise constant acceleration from a velocity-time graph?
Hint
Look for a constant slope over the interval.
Worked solution
The graph must be a straight line over that interval, including the horizontal case. Two endpoint readings alone do not prove the intermediate graph is straight.
07 / Separate a change of gradient from a jump in velocity
Idealised graph corners have limits.
13 · Joined straight stages
Two velocity-time segments meet continuously with different gradients. What changes at their join?
Hint
Height is continuous; slope changes.
Worked solution
Velocity remains continuous but acceleration changes. At the sharp corner the ordinary derivative is not uniquely defined; use the appropriate one-sided acceleration for each stage.
14 · Parallel graphs
Can two different straight velocity-time graphs have the same acceleration?
Hint
Their intercepts can differ.
Worked solution
Yes. Parallel lines have equal slopes and therefore equal acceleration, while their velocity values differ at the same time.
A vertical jump in velocity would represent an idealised instantaneous change, such as an impact, rather than a finite constant acceleration over zero time. Keep that event separate from the straight-line flight stages.
08 / Height, slope and signs answer different questions
Use the axes and interval deliberately.
Read signed velocity from the vertical coordinate. Calculate acceleration as the slope, with units m/s². Compare velocity and acceleration directions for speed change; split at reversals. A secant gives average acceleration, while a smooth tangent gives instantaneous acceleration.
Section 1 of 8 · Read velocity from height and acceleration from slope