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Durham College · PHYS 1131

Durham PHYS 1131 tutoring

Help understanding motion, forces and energy problems in Durham PHYS 1131.

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Course overview

PHYS 1131 covers motion, dynamics, statics, work and energy, torque and circular motion, electricity, fluids and heat transfer. Start with the chapter and physical system you are studying. The practice below shows how motion data can be used to find acceleration and force.

PHYS 1131 topics to review

Kinematics, dynamics and staticsWork, energy, torque and circular motionElectricity, fluids and heat transfer

Common difficulties

  • Using km/h in an equation expressed in SI units
  • Interpreting negative acceleration without specifying an axis

Preparing for this course

Review unit conversions, vector signs and the distinction between a force and an acceleration. Convert speed into metres per second before using SI formulas. In braking problems, choose one positive axis and retain it even when the acceleration points the other way.

Registration requirements

The guidance above is study preparation, not an eligibility decision. Confirm current prerequisites, restrictions and availability with Durham College.

Check the official course catalogue

Connect stopping distance, acceleration and force

A 900 kg vehicle moves at 72 km/h and stops uniformly in 50 m on a level road. Find its acceleration and the net horizontal braking force. Treat it as a particle.

Show the reasoning

  1. Choose units and signs

    Let mm be mass, aa acceleration, viv_i and vfv_f initial and final velocity, Δx\Delta x displacement and FnetF_{\mathrm{net}} net horizontal force. Take forward as positive. The initial speed is 72/3.6=20 m/s72/3.6=20\text{ m/s}, final speed is zero and displacement is +50 m+50\text{ m}.

    Why divide by 3.6?

    A conversion changes the unit, not the speed. Since 1 km=1000 m1\,\mathrm{km}=1000\,\mathrm m and 1 h=3600 s1\,\mathrm h=3600\,\mathrm s, multiply by ratios of equal quantities:

    72 kmh×1000 m1 km=72000 m/h.72\,\frac{\mathrm{km}}{\mathrm h}\times\frac{1000\,\mathrm m}{1\,\mathrm{km}}=72000\,\mathrm{m/h}.

    72000 mh×1 h3600 s=20 m/s.72000\,\frac{\mathrm m}{\mathrm h}\times\frac{1\,\mathrm h}{3600\,\mathrm s}=20\,\mathrm{m/s}.

    The kilometre and hour units cancel, leaving metres per second. Each ratio equals one, so the physical quantity is unchanged. The numerical factor 1000/36001000/3600 equals 1/3.61/3.6. Reverse the conversion by multiplying a speed in m/s by 3.6.

    Try the conversion

    An object travels at a constant speed of 54 km/h for 8.0 s. Convert its speed to m/s, then find the distance travelled.

    Check the answer

    54×1000/3600=15 m/s54\times1000/3600=15\,\mathrm{m/s}. Distance equals speed multiplied by time: (15 m/s)(8.0 s)=120 m(15\,\mathrm{m/s})(8.0\,\mathrm s)=120\,\mathrm m. Multiplying 54 by 8 directly mixes hours and seconds.

  2. Use the constant-acceleration model

    vf2=vi2+2aΔxv_f^2=v_i^2+2a\Delta x gives 0=400+100a0=400+100a, so a=−4 m/s2a=-4\text{ m/s}^2. The negative sign indicates acceleration opposite the forward axis.

  3. Find force and check work

    Fnet=ma=−3600 NF_{\mathrm{net}}=ma=-3600\text{ N}. Its work over 50 m is −180,000 J, equal to the change from 12(900)(202)=180,000 J\tfrac12(900)(20^2)=180{,}000\text{ J} to zero kinetic energy.

Try a changed problem

With the same acceleration magnitude, what stopping distance is needed from 36 km/h?

Check your reasoning

The initial speed is 10 m/s, so Δx=102/(2⋅4)=12.5 m\Delta x=10^2/(2\cdot4)=12.5\text{ m}. Halving speed quarters the stopping distance under this model.

How tutoring can help

Share a question, even if you have not started it. For mechanics, a tutor can help you connect motion data to a force diagram and work through a kinematics or energy solution. For electricity, fluids or heat transfer, tell us the topic so we can confirm whether we can help.

Exam preparation for PHYS 1131

Practise stating assumptions, selecting the system and converting units before calculation. Mix short conceptual checks with quantitative problems. Use the chapter list for your assessment to plan practice across the required physical-science topics.

Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Independent practice

Free practice and study guides for PHYS 1131

90 min

Work, Energy or Momentum: Choose the Method

Choose a system and interval to connect work, energy and momentum. Guided collision and force–position practice, with an optional centre-of-mass and relative-displacement extension.

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PHYS 1131 · official course reference