01 · Same signs
A trolley has v=+4 m/s and a=+2 m/s² on a right-positive axis. Describe its motion and local speed change.
Hint
Both vectors point right.
Worked solution
It moves right and speeds up locally.
Understand · explore · practise
Signs of velocity and acceleration
Use a chosen axis to distinguish direction of motion from speeding up or slowing down, including zero velocity and reversal.
Before you startScalars and vectors in mechanics; signed numbers.
01 / Velocity gives direction; acceleration changes velocity
Negative acceleration is not a synonym for slowing down.
For nonzero one-dimensional velocity, compare the signs of v and a.
Same nonzero signs mean speed increases locally; opposite signs mean speed decreases locally. This describes an instant or an interval on which those signs remain unchanged.
Direction and speed changeExplore
Right is positive. Each case gives instantaneous signed velocity v and acceleration a. For nonzero velocity, compare their directions to decide the local speed change.
02 / Start with motion in the positive direction
Here speed equals the positive velocity component.
02 · Opposite signs
Change the acceleration to −2 m/s² while v=+4 m/s. Describe the motion.
Hint
Acceleration points against the velocity.
Worked solution
It still moves right at this instant, but its speed decreases locally. Acceleration direction is not its current motion direction.
03 / Repeat the reasoning for negative velocity
A more negative velocity can have greater magnitude.
Watch: negative velocity becomes larger in magnitude
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Negative pair
A trolley has v=−4 m/s and a=−2 m/s². Is it slowing down?
Hint
Compare velocity magnitudes as the component becomes more negative.
Worked solution
No. It moves left and speeds up locally. For example, changing from −4 to −6 m/s increases speed from 4 to 6 m/s.
04 · Opposing acceleration
A trolley has v=−4 m/s and a=+2 m/s². Describe its motion.
Hint
The acceleration opposes the leftward velocity.
Worked solution
It moves left but slows down locally, until any later stopping or reversal changes the signs.
04 / An interval can include a reversal
Check whether velocity crosses zero before describing the whole interval.
Let v=−4+2t m/s for 0≤t≤3 s, with right positive.Worked example
a=+2 m/s² throughout
The signed acceleration is positive.
v=0 at t=2 s
Before this time velocity is negative; afterwards it is positive.
Speed falls from 4 to 0, then rises to 2 m/s
The same positive acceleration first slows the leftward motion and then speeds up rightward motion.
05 · One second
Find v and speed at t=1 s.
Hint
Substitute and take the magnitude.
Worked solution
v=−2 m/s; speed=2 m/s, moving left.
06 · Three seconds
Find v and speed at t=3 s.
Hint
The turning time has passed.
Worked solution
v=+2 m/s; speed=2 m/s, moving right. Equal speeds at t=1 and t=3 do not mean equal velocities.
05 / Zero velocity needs care
It does not automatically imply zero acceleration.
07 · At the turn
For v=−4+2t, state velocity and acceleration at t=2 s.
Hint
Differentiate or use the constant rate of change.
Worked solution
v=0 and a=+2 m/s². The particle is instantaneously at rest but not in a state of zero acceleration.
08 · Local rule at zero
Why not mechanically apply the same-sign/opposite-sign rule when v=0?
Hint
Zero has neither positive nor negative direction.
Worked solution
That rule assumed nonzero velocity. Examine the motion immediately before and after, using the model and acceleration behaviour.
09 · Zero acceleration
If v=−4 m/s and a=0 at an instant, what can you say at that instant?
Hint
Distinguish an instantaneous value from behaviour over a whole interval.
Worked solution
The particle moves left and its instantaneous acceleration is zero. If a remains zero throughout an interval, its velocity stays constant there; a single zero value does not establish that whole-interval claim.
06 / Location is not direction of travel
A particle can be left of the origin and move right.
10 · Separate quantities
A particle has x=−8 m and v=+3 m/s. Interpret both on a right-positive axis.
Hint
Use the origin for x and the axis for v.
Worked solution
It is 8 m left of the origin and moving right at 3 m/s.
11 · Origin crossing
Does crossing x=0 require velocity or acceleration to be zero?
Hint
The origin is a chosen reference point.
Worked solution
No. A particle can pass the origin with nonzero velocity and acceleration. A coordinate origin is not automatically a physical stopping point.
07 / State the time scope of a conclusion
A change over a finite interval may hide intermediate behaviour.
12 · Velocity change
Velocity changes from −7 to −3 m/s in 2 s. Find average acceleration and compare initial and final speeds.
Hint
Use final minus initial, divided by elapsed time.
Worked solution
Average acceleration=(−3−(−7))/2=+2 m/s². Initial speed 7 m/s exceeds final speed 3 m/s. These endpoints alone do not prove speed decreased at every intermediate instant.
13 · Axis reversal
If you reverse the axis, do the conclusions speeding up/slowing down change?
Hint
Both signed components reverse.
Worked solution
No. Both v and a change sign, preserving whether their directions agree or oppose. Speed is independent of the coordinate orientation.
14 · Wording
Rewrite “negative acceleration means slowing down” accurately for one-dimensional nonzero motion.
Hint
Mention the velocity direction.
Worked solution
Speed decreases locally when acceleration and velocity have opposite signs. Negative acceleration slows positive-direction motion but speeds negative-direction motion.
08 / Separate three questions
Where is it, which way is it moving, and how is its speed changing?
Position locates the particle; velocity gives direction and speed; acceleration changes velocity. For nonzero one-dimensional motion, compare velocity and acceleration signs. At a stop or across a reversal, inspect the model before and after the instant.
Section 1 of 8 · Velocity gives direction; acceleration changes velocity