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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 2.10.7 Linear Motion in Real-Life

Learner Experience 2.10.16.

Work in Groups
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The local government plans to introduce a new school bus stop along a straight road near your school. Before approval, the transport committee needs to determine whether the stopping distance is safe for learners crossing the road.
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You are part of a team of student consultants tasked with analysing the motion of a school bus using principles of linear motion.
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Your team is given the following data:
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  • The bus is travelling at \(20\) m/s along a straight road.
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  • The driver notices learners crossing \(50\) metres ahead.
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  • The driver’s reaction time is \(2\) seconds.
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  • After reacting, the bus decelerates uniformly at \(4\) m/s\(^2\) until it stops.
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What to do
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(a)

Reaction distance is the distance the vehicle travels from the moment the driver perceives a hazard to the moment they apply the brakes.
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  1. Calculate the distance travelled during the \(2\)-second reaction time.
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  2. State whether the bus is accelerating, decelerating, or moving at constant velocity during this time.
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(b)

Braking distance is the distance it takes for the bus to come to a complete stop after the driver starts braking.
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  1. Calculate the time taken for the bus to come to rest after braking begins.
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  2. Calculate the braking distance using the formula \(v^2 = u^2 + 2as\text{.}\)
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(c)

Total stopping distance is the sum of the reaction distance and the braking distance.
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  1. Calculate the total stopping distance of the bus.
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  2. Compare this with the \(50\) m available before the crossing.
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  3. Conclude whether the bus will stop in time.
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(d)

Find:
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  1. The total displacement of the bus from the moment the driver notices the learners to when the bus stops.
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  2. The average velocity during the entire stopping process.
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(e)

Based on your calculations, recommend one of the following:
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  1. Reduce the speed limit
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  2. Move the bus stop further away
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  3. Install a speed bump
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  4. Maintain the current arrangement
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(f)

Share your work with other learners
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Key Takeaway 2.10.64.

Linear motion principles are fundamental in transportation safety.
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Understanding stopping distances helps engineers and planners design safer roads and establish appropriate speed limits in different zones.
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The relationship between speed, reaction time, and braking distance directly impacts pedestrian safety and accident prevention.
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Engineers apply motion calculations when designing vehicles, elevators, and transport systems to ensure safety and efficiency.
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In sports, motion analysis improves performance by evaluating speed, acceleration, and timing.
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Industries use constant motion principles to control machines and production systems.
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Example 2.10.65.

A car is travelling at a velocity of \(20\) m/s. The driver applies brakes and the car comes to rest in \(5\) seconds.
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  1. Calculate the acceleration of the car.
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  2. Calculate the stopping distance.
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  3. Explain why higher speeds increase accident risk.
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Solution.
  1. u = \(20\) m/s\(^1\)
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    v = \(0\)m/s \(^1\)
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    t = \(5\) s
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    a = \(\frac{-20}{5}\)
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    a = \(-4\) m/s\(^2\)
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  2. s = ut + \(\frac{1}{2}\)at\(^2\)
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    s = \(50\) m
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  3. When the initial speed increases, the stopping distance increases. A higher velocity means the vehicle covers more distance before it can come to rest, increasing the likelihood of a collision.
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Example 2.10.66.

A sprinter starts from rest and accelerates uniformly at \(3\) m/s\(^2\) for \(4\) seconds.
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  1. Calculate the final velocity.
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  2. Calculate the distance covered during this time.
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  3. Explain how improving acceleration benefits performance.
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Solution.
  1. v = u + at
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    v \(= 0 + 3 \times 4\)
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    \(= 12 \)m/s
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  2. Greater acceleration allows an athlete to reach a higher speed in a shorter time. This provides a competitive advantage, especially at the start of a race.
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Example 2.10.67.

An elevator accelerates uniformly from rest to \(4\) m/s in \(2\) seconds.
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  1. Calculate the acceleration.
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  2. Explain why gradual acceleration is important.
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Solution.
  1. a = (v - u) / t
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    a \(= \frac{(4 - 0)} {2}\)
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    a\(= 2\) m/s\(^2\)
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  2. Gradual acceleration ensures passenger comfort and safety. Sudden acceleration would cause passengers to lose balance and could result in injury.
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Checkpoint 2.10.68.

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Exercises Exercises

1.

A student walks a distance of \(10\) m in \(8 \) s. Another student runs the same distance in \(3\) s.
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  1. Calculate the walking speed.
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  2. Calculate the running speed.
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  3. State how a Physical Education teacher could use this information to plan training.
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Answer.
  1. \(1.25\) m/s
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  2. \(3.33\) m/s
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  3. The teacher can set realistic speed targets and design drills to improve students’ acceleration and endurance.
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2.

A car travels at a constant velocity of \(20\) m/s. The driver has a reaction time of \(1\) s.
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  1. Calculate the distance travelled during the driver’s reaction time.
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  2. Explain why lower speed limits are enforced in pedestrian areas.
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Answer.
  1. Reaction distance = \(20 \times 1 = 20\) m
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  2. Lower speeds reduce both reaction distance and braking distance, giving drivers more time to stop and improving pedestrian safety.
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3.

Two cars are travelling at \(25\) m/s.
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Car A comes to rest in \(5\) s.
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Car B comes to rest in \(8\) s.
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  1. Calculate the acceleration of each car.
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  2. Determine which car has the shorter stopping distance.
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  3. Explain which braking system is more effective.
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Answer.
  1. Car A \(-5 \, \text{m/s}^2\)
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    Car B \(-3.125 \, \text{m/s}^2\)
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  2. Car A has the shorter stopping distance.
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  3. Car A’s braking system is more effective because it produces a larger deceleration. A greater negative acceleration reduces stopping distance, which improves road safety.
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4.

A cyclist travelling at \(5\) m/s takes \(2\) s to react and brake to a stop. A child runs into the road \(12\) m ahead.
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  1. State the equation relating distance, speed and time.
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  2. Calculate the cyclist’s reaction distance.
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  3. Determine and explain whether the cyclist will stop before reaching the child.
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Answer.
  1. Distance = speed \(\times\) time
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  2. Reaction distance = \(5 \times 2 = 10\)
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  3. Yes, the cyclist will stop before reaching the child.
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    The reaction distance of \(10\) m is less than the \(12\) m available, leaving a safety margin of \(2\) m
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