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TestAS · Engineering

Physics: TestAS practice testlets

Motion, forces, energy, pressure and flow. A text explains the law, for example how stopping distance grows with speed, and the questions ask you to calculate with it.

testas.de lists this topic for the Engineering module. In the test it comes as testlets: a short text, table or formula, then 4 to 8 questions you answer from it.

Topic list: testas.de

Two practice testlets

Full exam-length testlets from the Kollego course, generated the same way as our mocks. Answer from the material, then open each answer to check.

Testlet 1

Acceleration and stopping distance of a car

Kinematics describes motion without asking about its causes. The velocity v of a body is the rate of change of its position, and the acceleration a is the rate of change of its velocity. For motion with constant acceleration starting from rest, the velocity after a time t is v = a·t and the distance travelled is s = ½·a·t². In a velocity-time graph, the slope is the acceleration and the area under the graph is the distance travelled. Speeds in traffic are usually given in km/h; to convert into m/s, divide by 3.6, so 108 km/h equals 30 m/s.

v = a·t s = ½·a·t² braking distance = v² ÷ (2·a) 1 m/s = 3.6 km/h

When a driver sees a hazard, the car does not begin to slow down at once. During the reaction time, typically about one second, the car continues at constant speed; the reaction distance is speed times reaction time. Then the brakes act, and the car decelerates until it stops. For a constant deceleration a, the braking distance is v² divided by 2a. The stopping distance is the sum of reaction distance and braking distance.

The square in the formula for the braking distance has important consequences for road safety. If the speed doubles, the reaction distance doubles, but the braking distance quadruples. On a wet road, the maximum deceleration falls, for example from about 8 m/s² to about 5 m/s², so the braking distance becomes much longer. Speed limits near schools and in residential areas are based on these relationships.

Real cars do not accelerate perfectly uniformly: electric motors deliver high torque from standstill, while combustion engines need gear changes. For planning and safety calculations, however, the model of constant acceleration gives useful estimates. Driver assistance systems use the same equations: an emergency brake assistant calculates from the measured distance and relative speed whether a collision can still be avoided and brakes automatically if not.

Engineers testing a new electric car measured its velocity during a standing start and during an emergency stop on a dry test track. The results are shown below. In the test, the car accelerated uniformly and braked with a constant deceleration of 6 m/s². The test driver's reaction time was measured at 1.0 s.

Velocity during the standing start
Time (s)0246810
Velocity (m/s)0612182430
  1. 1. What is the car's acceleration during the standing start?

    • a6.0 m/s²
    • b0.3 m/s²
    • c3.0 m/s²
    • d30.0 m/s²
    Show answer

    C: 3.0 m/s²

    30 m/s in 10 s.

  2. 2. How far does the car travel in the first 10 seconds?

    • a75 m
    • b150 m
    • c300 m
    • d30 m
    Show answer

    B: 150 m

    ½ × 3 × 10².

  3. 3. What speed in km/h does the car reach after 10 s?

    • a83 km/h
    • b30 km/h
    • c180 km/h
    • d108 km/h
    Show answer

    D: 108 km/h

    30 × 3.6.

  4. 4. What is the braking distance from 30 m/s at 6 m/s²?

    • a150.0 m
    • b37.5 m
    • c5.0 m
    • d75.0 m
    Show answer

    D: 75.0 m

    30² ÷ (2 × 6) = 900 ÷ 12.

  5. 5. What is the total stopping distance from 30 m/s, including the reaction time?

    • a135 m
    • b105 m
    • c80 m
    • d75 m
    Show answer

    B: 105 m

    30 m reaction distance + 75 m braking distance.

  6. 6. From 15 m/s, how does the braking distance compare with that from 30 m/s?

    • aIt is one eighth as long
    • bIt is the same length
    • cIt is half as long as before
    • dIt is one quarter as long
    Show answer

    D: It is one quarter as long

    Braking distance ∝ v².

Testlet 2

Work, energy, power and efficiency of a crane

In physics, work is done when a force moves its point of application: W = F·s, if the force acts along the direction of motion. Work is measured in joules (J); one joule is one newton times one metre. Energy is the capacity to do work. Lifting a mass m by a height h against gravity requires work equal to the gain in potential energy, E_pot = m·g·h. A moving mass has kinetic energy E_kin = ½·m·v².

W = F·s E_pot = m·g·h E_kin = ½·m·v² P = W ÷ t η = P_out ÷ P_in

Energy is conserved: it can be converted between forms but not created or destroyed. If friction and air resistance are neglected, a falling body converts potential into kinetic energy, so that m·g·h = ½·m·v², and its speed after falling a height h is v = √(2·g·h), independent of its mass.

Power is the rate at which work is done or energy converted: P = W ÷ t, measured in watts (1 W = 1 J/s). A machine that does the same work in half the time has twice the power. Because every real machine loses some energy, mostly as heat through friction and electrical resistance, its output power is smaller than its input power. The efficiency η is the ratio of useful output to input, often given in percent.

Kinetic energy grows with the square of the speed. A car at twice the speed has four times the kinetic energy, which must be converted into heat by the brakes when it stops, or into deformation in a collision. Energy is also measured in kilowatt-hours on electricity bills: 1 kWh = 3.6 million joules.

In the test, some of the electrical input was converted into heat in the motor windings and into friction in the gearbox and the rope pulleys, which explains why the output was smaller than the input.

A construction crane with an electric motor was tested on site. In the test it lifted a pallet of bricks at constant speed. Energy for the lift came from the mains, and the motor's electrical input power was measured during the lift.

Crane test (g = 10 m/s²)
QuantityValue
Mass of pallet500 kg
Height lifted20 m
Time for the lift25 s
Electrical input power of the motor5 kW
  1. 1. How much potential energy does the pallet gain?

    • a250 kJ
    • b1,000 kJ
    • c100 kJ
    • d10 kJ
    Show answer

    C: 100 kJ

    500 × 10 × 20 J.

  2. 2. What useful output power does the crane deliver?

    • a100.0 kW
    • b4.0 kW
    • c5.0 kW
    • d2.5 kW
    Show answer

    B: 4.0 kW

    100,000 J ÷ 25 s.

  3. 3. What is the efficiency of the crane?

    • a20%
    • b125%
    • c40%
    • d80%
    Show answer

    D: 80%

    4 kW ÷ 5 kW.

  4. 4. If the pallet were dropped from 20 m, with what speed would it hit the ground, ignoring air resistance?

    • a200 m/s
    • b400 m/s
    • c14 m/s
    • d20 m/s
    Show answer

    D: 20 m/s

    √(2 × 10 × 20) = √400.

  5. 5. A car of 1,000 kg travels at 20 m/s. What is its kinetic energy?

    • a10 kJ
    • b200 kJ
    • c400 kJ
    • d20 kJ
    Show answer

    B: 200 kJ

    ½ × 1,000 × 20².

  6. 6. How much electrical energy does the motor use for the lift?

    • a25 kJ
    • b5 kJ
    • c100 kJ
    • d125 kJ
    Show answer

    D: 125 kJ

    5,000 W × 25 s.

How to work a testlet

Other Engineering topics

Mathematics and statisticsChemistryMaterialsTechnical mechanicsThermodynamicsElectrical engineeringComputer scienceDevelopment and construction

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