01 · No acceleration needed
A particle changes uniformly from 4 to 10 m/s in 6 s. Find displacement.
Hint
Use the equation omitting a.
Worked solution
s = ½(u + v)t = ½(4 + 10) × 6 = 42 m.
Understand · explore · practise
Choosing and using SUVAT equations
Choose a constant-acceleration equation from known and required quantities, convert units, keep signs and check physical answers.
Before you startThe five SUVAT equations; rearranging and unit conversion.
01 / Write the known values and the target
Choose an equation that avoids an unnecessary unknown.
Start with a signed quantities table.
State the positive direction, identify one constant-acceleration interval, convert units, then list s, u, v, a and t. Mark the requested quantity. The unused quantity guides equation choice.
Which variable can you leave out?Explore
Choose the quantity that is neither supplied nor requested. The displayed equation excludes it. First verify that acceleration is constant and the given values refer to one interval.
02 / Match the missing variable
Do not select solely because a formula looks familiar.
A particle starts at 3 m/s and accelerates at 2 m/s² for 5 s. Find its final velocity.Worked example
u = 3, a = 2, t = 5; target v
Displacement s is unnecessary.
v = u + at
This equation leaves out s.
v = 3 + 2 × 5 = 13 m/s
The units and positive direction are consistent.
02 · No final velocity needed
A particle starts at 2 m/s with a = 3 m/s² for 4 s. Find displacement.
Hint
Use s = ut + ½at².
Worked solution
s = 2 × 4 + ½ × 3 × 16 = 32 m.
03 / Convert before substituting
A kilometre-per-hour value cannot be mixed directly with metres and seconds.
Watch: convert the speed before calculating
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Kilometres per hour
A car starts at 54 km/h and accelerates at 1.5 m/s² for 8 s. Find final speed in m/s.
Hint
Divide 54 by 3.6.
Worked solution
54 km/h = 15 m/s. Then v = 15 + 1.5 × 8 = 27 m/s.
04 · Minutes
A particle starts from rest at a = 0.2 m/s² for 1.5 minutes. Find displacement.
Hint
Use t = 90 s.
Worked solution
s = ½ × 0.2 × 90² = 810 m.
04 / Keep velocities and acceleration signed
Leftward motion can have rightward acceleration.
05 · Leftward initial velocity
Right is positive. A particle starts leftward at 7 m/s and accelerates rightward at 2 m/s² for 5 s. Find final velocity and displacement.
Hint
Use u = −7 and a = +2.
Worked solution
v = −7 + 2 × 5 = +3 m/s. s = −7 × 5 + ½ × 2 × 25 = −10 m. The final rightward velocity does not make the total displacement positive.
06 · Braking
A vehicle moves at 18 m/s and stops uniformly in 6 s. Find its signed acceleration and stopping distance, with forward positive.
Hint
Set v = 0.
Worked solution
a = (0 − 18)/6 = −3 m/s². s = ½(18 + 0) × 6 = 54 m. Velocity stays nonnegative until rest, so displacement equals distance.
05 / Rearrange carefully when the target is not isolated
Show the equation before inserting a calculator result.
07 · Find acceleration
A particle starts at 5 m/s, covers 44 m in 4 s, and has constant acceleration. Find a.
Hint
44 = 5 × 4 + ½a × 4².
Worked solution
44 = 20 + 8a, so a = 3 m/s².
08 · Find initial velocity
A particle ends at 9 m/s after 5 s and covers 30 m with constant acceleration. Find u.
Hint
Use s = ½(u + v)t.
Worked solution
30 = ½(u + 9) × 5, so u + 9 = 12 and u = 3 m/s.
09 · Omit initial velocity
Given v = 8 m/s, a = 1 m/s² and t = 6 s, find displacement.
Hint
Use s = vt − ½at².
Worked solution
s = 8 × 6 − ½ × 1 × 36 = 30 m.
06 / A squared velocity still needs a direction check
A speed is nonnegative; a velocity can be negative.
10 · No time supplied
A particle starts at 4 m/s and accelerates at 2 m/s² over 21 m in the positive direction. Find final velocity.
Hint
Use v² = u² + 2as, then check motion direction.
Worked solution
v² = 16 + 84 = 100. Since u > 0 and a > 0 for forward elapsed time, v = +10 m/s.
11 · Decelerating but still forward
A vehicle slows from 20 to 8 m/s with a = −4 m/s². Find the distance covered.
Hint
Rearrange the squared equation for s.
Worked solution
s = (8² − 20²)/(2 × (−4)) = 42 m. The vehicle stays forward-moving, so this is also distance.
07 / Reject calculations outside the model
A neat number alone is not evidence that a formula applies.
12 · Inconsistent braking data
A forward-moving car is said to go from 12 m/s to rest while its signed acceleration is +3 m/s². What does v = u + at reveal?
Hint
Solve 0 = 12 + 3t.
Worked solution
It gives t = −4 s, inconsistent with a future stopping event. With forward positive, uniform braking requires negative acceleration.
13 · Acceleration varies
A velocity graph is curved over an interval. Can you use the five constant-acceleration equations for that entire interval?
Hint
A curved velocity graph has a changing slope.
Worked solution
Not in general. Use the actual graph or a suitable variable-acceleration method; only use a constant approximation if it is explicitly justified.
14 · Different intervals
One speed is measured before a stop and another after restarting. May those values be inserted into one SUVAT calculation?
Hint
Check whether one constant acceleration describes everything between them.
Worked solution
Usually no. Separate the stopping, stationary and restarting stages, using their own initial/final velocities and durations.
08 / Choose, substitute, solve and interpret
An equation choice is part of a model, not a substitute for it.
Identify a constant-acceleration interval and signed axis. Convert units, list the knowns and target, choose the equation omitting the unused variable, then solve. Check time, direction and units, and split a reversal before finding total distance.
Section 1 of 8 · Write the known values and the target