Kinematics in a straight line flashcards

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Displacement
The position of a particle relative to a fixed point (or its start), measured along the line with a sign. A vector; unit m.
Distance travelled
The total length of path covered. A scalar that never decreases, even when the particle turns back.
Velocity (in one dimension)
The rate of change of displacement, with a sign showing the direction. Unit m s⁻¹.
Speed
The magnitude of the velocity: always positive or zero.
Acceleration
The rate of change of velocity, with a sign. Unit m s⁻². A deceleration is an acceleration opposite to the velocity.
Average speed and average velocity
Average speed = total distance ÷ total time; average velocity = total displacement ÷ total time.
Moving freely under gravity
Moving with only the weight acting (no air resistance): constant acceleration g = 9.8 m s⁻² vertically downwards.
Constant acceleration: the formula without s
v = u + at (No — must be known)
Constant acceleration: the formula without a
s = ½(u + v)t (No — must be known)
Constant acceleration: the formula without t
v² = u² + 2as (No — must be known)
Constant acceleration: the formula without v
s = ut + ½at² (No — must be known)
Constant acceleration: the formula without u
s = vt − ½at² (No — must be known)
Greatest height reached by a particle projected vertically upwards at u
H = u²/(2g), reached after u/g seconds (from v = 0 at the top). (No — derive it)
Gradient of a displacement–time graph
The velocity. Horizontal line = at rest; straight line = constant velocity; turning point = instantaneously at rest. (No — must be known)
Gradient and area of a velocity–time graph
Gradient = acceleration; area between the graph and the t-axis = displacement (below the axis counts as negative). (No — must be known)
Area under a speed–time graph
The distance travelled. The speed–time graph is the v–t graph with the parts below the axis reflected above it. (No — must be known)
Area under an acceleration–time graph
The change in velocity over that interval (add the initial velocity to get the velocity). (No — must be known)
Shape of the v–t graph for accelerate – steady speed – decelerate
A trapezium: area = ½ × (time at steady speed + total time) × top speed.
v–t graph of a ball thrown upwards and caught at the same level (up positive)
A straight line of gradient −9.8 from +u to −u, crossing the t-axis at the top of the flight; equal areas above and below the axis.
Why can't the suvat formulae be used across a whole multi-stage journey?
They need one constant acceleration. Split the motion where the acceleration changes; the final velocity of one stage is the initial velocity of the next.
Acceleration at the highest point of a vertical throw
Still 9.8 m s⁻² downwards — only the velocity is zero there.
When does one vehicle overtake (or meet) another?
When their displacements from the same origin, at the same time, are equal (equal areas under their v–t graphs).
When is the gap between two vehicles greatest before one catches the other?
When their velocities are equal — where their v–t graphs cross.
Why can s = ut + ½at² give two positive times for the same displacement?
If u and a have opposite signs the particle passes the point, stops, turns back and passes it again.
Modelling assumptions for vertical motion under gravity
The body is a particle (size ignored), air resistance is ignored, and g is constant at 9.8 m s⁻².
A stone is dropped from rest and falls 10 m. Speed on landing?
v² = 0 + 2 × 9.8 × 10 = 196, so v = 14 m s⁻¹.
A car at 25 m s⁻¹ brakes with deceleration 5 m s⁻². Stopping distance?
0 = 25² − 2 × 5 × s ⇒ s = 62.5 m.
A cart decelerates at 3 m s⁻² and stops after 6 s. Distance in those 6 s?
s = vt − ½at² with v = 0, a = −3: s = 0 + ½ × 3 × 36 = 54 m.
A ball is projected upwards at 19.6 m s⁻¹ from the ground. Time of flight and greatest height?
Time = 2u/g = 4 s; height = u²/(2g) = 19.6 m. (No — derive it)
A v–t trapezium: top speed 15 m s⁻¹, steady for 40 s, total time 64 s. Distance?
½(40 + 64) × 15 = 780 m.
A car starts from rest at 0.5 m s⁻² as a cyclist passes at a constant 5 m s⁻¹. When does the car catch up?
½ × 0.5t² = 5t ⇒ t = 20 s (the t = 0 root is the start).
v = 6 − 2t for 0 ≤ t ≤ 4. Displacement and distance?
Displacement = 6 × 4 − 4² = 8 m. Stops at t = 3 after 9 m, then 1 m back: distance = 10 m.
Signs in vertical motion
Fix a positive direction first. With up positive, a = −9.8 and a point below the start has negative s.
Choosing a suvat formula
List s, u, v, a, t; mark the three known and the one wanted; use the formula that leaves out the fifth.
A quadratic gives two times
Test each root: reject negative times and any root outside the stage the formula describes — and say why.
Area under a v–t graph partly below the axis
Displacement = area above − area below; distance = area above + area below.
Accuracy with g = 9.8
Give final answers to 2 or 3 significant figures; keep unrounded values between parts.
Object released from a moving balloon or lift
Its initial velocity is the velocity of the balloon or lift at release, not zero.

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