Laws of Motion · Lesson 13 of 14
Changing Forces, Motion Graphs and Multi-Stage Problems
“Link force intervals to acceleration, velocity changes and impulses without extending a model beyond its stated duration.”
• Solve motion before, during and after a finite-duration force. • Infer acceleration and impulse from a piecewise position-time graph. • Distinguish time-averaged and distance-averaged resistance. • Check the mass and system included in a force calculation.
One force model may describe only one phase
A force can turn on, remain constant for a specified duration, then turn off. Acceleration changes between phases, while position and usually velocity remain continuous across the boundaries. Write a separate equation for each phase and carry the final state into the next. The clock can continue through all phases, but each local equation must use time measured from the start of that phase.
Problem
A 0.40 kg body initially moves north at 10 m/s. An 8 N southward force acts from t = 0 to 30 s, and no other net force acts before or afterwards. Take north positive and x(0) = 0. Find positions at −5, 25 and 100 s.
- 1.Before t = 0, velocity is +10 m/s. Thus x(−5) = 10 × (−5) = −50 m.
- 2.While the force acts, a = −8/0.40 = −20 m/s², so x = 10t − 10t² and v = 10 − 20t.
- 3.At 25 s, x = 250 − 6250 = −6000 m. The body has already reversed because v becomes zero at 0.5 s.
- 4.At 30 s, x = 300 − 9000 = −8700 m and v = 10 − 600 = −590 m/s.
- 5.After 30 s, the force is off and velocity stays −590 m/s. At 100 s, x = −8700 − 590(100 − 30) = −50,000 m. Continuing the accelerated equation to 100 s would apply a force outside its given interval.
A constant retarding force and a continuing fixed-direction force differ
A braking force stated to act until rest ends its slowing phase at zero speed. A fixed southward force stated to act for 30 s continues south even after a northward-moving body reverses. At that later stage it increases southward speed. Use the actual force specification instead of assuming that a force always follows the direction opposite the instantaneous velocity.
Read slopes before inferring forces from a graph
For one-dimensional motion, the slope of x–t is velocity. A straight segment means constant velocity, so force is zero on its open interval for a fixed-mass particle. A sharp change of slope represents an idealised velocity jump. Its finite impulse is m times the slope change; it cannot be described by one finite instantaneous force at the corner.
Problem
For the graph above, find force in each open interval and the impulses at 0 and 4 s.
- 1.Before 0 s, velocity is zero. From 0 to 4 s, velocity is (3 − 0)/(4 − 0) = +0.75 m/s. After 4 s, velocity returns to zero.
- 2.Velocity is constant within each open interval, so acceleration and net force are zero there.
- 3.At 0 s, impulse is 4(0.75 − 0) = +3 N s. At 4 s, impulse is 4(0 − 0.75) = −3 N s.
- 4.A real interaction rounds the corners over a short time. Contact duration or a force-time shape is needed to determine force magnitude; the idealised graph gives the impulses.
A stopping distance does not uniquely give a time average
Problem
A 0.040 kg projectile enters a heavy block at 90 m/s and stops over 0.60 m. Estimate the opposing force.
- 1.Under a constant-deceleration model, 0 = 90² + 2a × 0.60, giving a = −6750 m/s².
- 2.The corresponding force is ma = −270 N, so its magnitude is 270 N opposing entry.
- 3.The same 270 N is the distance-averaged resistive-force magnitude if resistance varies along the stopping path: the kinetic-energy loss ½mu² = 162 J divided by 0.60 m is 270 N.
- 4.A time-averaged force requires stopping time, or sufficient information about the force history. Under constant deceleration, stopping time is 90/6750 = 0.01333 s and Δp/Δt also gives 270 N. Without that assumption, a time average need not equal the distance average.
The source calls this an average resistive force after using a uniform-deceleration equation. The distinction above states precisely what is inferred from the supplied stopping distance. The general impulse theorem involves a time integral; the energy-work relation used for the distance average involves force integrated along displacement. They are different averages.
Include the whole accelerated mass
Problem
A three-wheeler of mass 400 kg and driver of mass 65 kg slows from 36 km/h to rest in 4 s. Find average external retarding force on the combined system.
- 1.Initial speed is 10 m/s. Average acceleration along the initial direction is (0 − 10)/4 = −2.5 m/s².
- 2.The mass of the selected vehicle-plus-driver system is 465 kg.
- 3.Average external horizontal force is 465 × (−2.5) = −1162.5 N, about 1.2 × 10³ N opposite the initial motion.
- 4.If the driver were omitted while the same acceleration were used, the calculation would describe a different system. The braking mechanism’s internal forces do not replace the external road interaction in the combined diagram.
Thrust must overcome weight as well as produce acceleration
Problem
A rocket has lift-off mass 20,000 kg and initial upward acceleration 5 m/s². Take g = 10 m/s² and neglect drag. Find initial thrust.
- 1.At the initial instant, the upward thrust Q and downward weight satisfy Q − mg = ma.
- 2.Q = m(g + a) = 20,000 × 15 = 300,000 N upward.
- 3.Using Q = ma alone would omit 200,000 N of weight. This is an instantaneous force balance with thrust treated as a given interaction force; it is not a derivation of variable-mass rocket motion.
Quiz
A position-time graph has a sharp change from slope 0 to slope 2 m/s for a 3 kg particle. What does it determine?
Key Takeaways
• Use separate phases when forces turn on or off. • Carry position and velocity continuously between ordinary phase boundaries. • Straight position-time segments imply zero acceleration in their interiors. • A sharp velocity jump specifies impulse, not a finite force without duration. • Distinguish time averages from distance averages and define the system mass.