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Combine driving force minus resistance equals ma with SUVAT, infer unknown forces and separate powered motion from braking stages.
Before you startNewton’s second law and constant-acceleration equations.
01 / Connect a force diagram with a motion equation
For rightward motion with backward resistance: D − R = ma.
Choose rightward positive. A positive result accelerates rightward; a negative result initially slows rightward motion. Constant driving force, resistance and mass give constant acceleration while that force model applies.
A 4 kg particle initially moves right at 6 m/s. Constant rightward pull D acts against 4 N resistance. Change D to see the velocity during the next 6 s. The D = 0 case reaches rest exactly at 6 s; the graph makes no claim about later motion.
02 / Calculate the resultant, then the new velocity
1500 − 300 = 600a, giving a = 2 m/s²
Resolve horizontally.
v = 4 + 2 × 5 = 14 m/s
Use constant acceleration for the stated interval.
s = 4 × 5 + ½ × 2 × 5² = 45 m
The velocity stays positive, so distance equals displacement.
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A 5 kg particle has 18 N rightward pull and 8 N resistance. It starts at 3 m/s right. Find acceleration and velocity after 4 s.
Use D − R = ma, then v = u + at.
a = (18 − 8)/5 = 2 m/s²; v = 3 + 2 × 4 = 11 m/s right.
Find the distance travelled in question 1.
Check that there is no reversal.
s = 3 × 4 + ½ × 2 × 4² = 28 m. Velocity stays positive, so this is also distance.
03 / Constant velocity means the horizontal forces balance
A vehicle moves with constant velocity against resistance 450 N. Find its driving force.
Zero acceleration gives zero horizontal resultant.
D − 450 = 0, so D = 450 N.
A 300 kg vehicle increases speed from 6 to 14 m/s in 4 s under driving force 900 N and constant resistance. Find the resistance.
First a = (v − u)/t.
a = 2 m/s², so 900 − R = 300 × 2 and R = 300 N.
04 / Switching off the drive creates a new stage
0 − 300 = 600a, so a = −0.5 m/s²
A new force equation gives a new acceleration.
0 = 14 − 0.5t gives t = 28 s
Use a new clock at switch-off.
0² = 14² + 2(−0.5)s gives s = 196 m
This is the coasting distance only.
A 4 kg particle moves right at 6 m/s. Its pull is removed and resistance is 4 N until it stops. Find stopping time.
The new acceleration is negative.
a = −4/4 = −1 m/s²; 0 = 6 − t, so t = 6 s.
Find the stopping distance in question 5.
Use v² = u² + 2as.
0 = 36 − 2s, giving s = 18 m.
05 / Include every stated backward force
A 600 kg vehicle moves at 14 m/s. Its drive is off, background resistance is 300 N and a further 3000 N braking force acts backward. Find acceleration.
Both forces oppose its current motion.
−300 − 3000 = 600a, so a = −5.5 m/s².
Find stopping distance under the forces in question 7.
Use the unrounded acceleration.
0 = 14² + 2(−5.5)s, so s = 196/11 = 17.8 m to 3 significant figures.
A 1000 kg vehicle must stop from 12 m/s in 24 m. Drive is off and background resistance is 500 N. Find the additional constant braking force B.
Find a from the stopping distance, then resolve forces.
0 = 144 + 48a gives a = −3 m/s². −B − 500 = 1000(−3), so B = 2500 N.
06 / Use distance and time to infer the force
A 400 kg vehicle travels 60 m in 4 s from initial speed 10 m/s. Resistance is 200 N and acceleration is constant. Find driving force.
Use s = ut + ½at² first.
60 = 40 + 8a gives a = 2.5 m/s². D − 200 = 400 × 2.5, so D = 1200 N.
A particle increases speed from 2 to 8 m/s in 3 s under pull 17 N and resistance 5 N. Find its mass.
Find acceleration before rearranging F = ma.
a = 2 m/s² and resultant is 12 N, so m = 12/2 = 6 kg.
07 / Stop each model when its assumptions change
In question 5, can v = 6 − t be used at t = 8 s to claim the particle moves left at 2 m/s?
Resistance was specified while it moved right and until rest.
No. At 6 s it stops; the model’s stated interval ends. Resistance opposing motion cannot simply remain leftward to cause reversal without another force model.
Find total time and distance for the 600 kg powered and coasting worked examples together.
Add complete stage durations and distances.
Total time = 5 + 28 = 33 s; distance = 45 + 196 = 241 m. One average acceleration must not replace the two stage equations.
On a smooth horizontal surface a particle has horizontal pull D and no other horizontal force. What replaces D − R = ma?
Smooth means no friction in this model.
D = ma. The vertical reaction still balances weight if there is no vertical acceleration and no other vertical force.
08 / Use one force model per motion stage
Resolve driving, braking and resistance forces; use the resulting acceleration only over the interval where those forces apply. Carry velocity between stages, restart the local clock, and distinguish a stage distance from the total journey.
Section 1 of 8 · Connect a force diagram with a motion equation