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The graph shows how the velocity of a small car changes with time - Edexcel - GCSE Physics - Question 3 - 2013 - Paper 1

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The graph shows how the velocity of a small car changes with time. (a) Complete the sentence by putting a cross (✗) in the box next to your answer. The resultant fo... show full transcript

Worked Solution & Example Answer:The graph shows how the velocity of a small car changes with time - Edexcel - GCSE Physics - Question 3 - 2013 - Paper 1

Step 1

Complete the sentence by putting a cross (✗) in the box next to your answer.

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Answer

The resultant force on the car will be zero when the car is D moving at a constant velocity. This is because when the velocity is constant, the acceleration is zero, hence the net force acting on the car is also zero according to Newton's first law of motion.

Step 2

Use the graph to estimate the velocity of the car at three seconds.

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Answer

At three seconds, the estimated velocity of the car is approximately 10 m/s.

Step 3

Calculate the acceleration of the car when it is speeding up.

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Answer

To calculate acceleration when the car is speeding up, we can use the formula:

ext{Acceleration} = rac{ ext{Change in Velocity}}{ ext{Change in Time}}

Assuming the car goes from 0 m/s to 20 m/s over 10 seconds, the acceleration is:

ext{Acceleration} = rac{20 ext{ m/s} - 0 ext{ m/s}}{10 ext{ s}} = 2 ext{ m/s²}

Step 4

Explain why the units of acceleration are m/s².

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Answer

The units of acceleration are m/s² because acceleration measures the rate of change of velocity per unit time. This means that for every second, the velocity is changing by a certain number of meters per second, hence the unit of acceleration combines the units of distance (meters) over time (seconds) squared.

Step 5

Show that the car travels further at a constant velocity than it does when it is slowing down.

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Answer

When the car travels at a constant velocity, let's assume it maintains 20 m/s for 3 seconds. The distance traveled can be calculated using:

extDistance=extVelocityimesextTime=20extm/simes3exts=60extm ext{Distance} = ext{Velocity} imes ext{Time} = 20 ext{ m/s} imes 3 ext{ s} = 60 ext{ m}

When slowing down, assuming it decelerates from 20 m/s to 0 m/s in 2 seconds, the average velocity during this time period is:

ext{Average Velocity} = rac{20 ext{ m/s} + 0 ext{ m/s}}{2} = 10 ext{ m/s}

Thus, the distance traveled while slowing down is:

extDistance=extAverageVelocityimesextTime=10extm/simes2exts=20extm ext{Distance} = ext{Average Velocity} imes ext{Time} = 10 ext{ m/s} imes 2 ext{ s} = 20 ext{ m}

Comparing the two distances:

  • Constant velocity: 60 m
  • Slowing down: 20 m

Therefore, the car travels further at a constant velocity than when it is slowing down.

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