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Analyze the motion of a toy car using ticker tape data, calculating initial velocity, final velocity, and acceleration.

Physics

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5. Ticker Tape Analysis

a) Describe the motion of the toy car.
The distances between consecutive dots on the ticker tape are increasing, while the time interval between dots is constant. This indicates that the velocity of the toy car is increasing. Therefore, the toy car is undergoing accelerated motion.

b) i) Calculate the initial velocity, u, of the toy car.

  1. Identify the distance and time for the first segment.

    The distance of the first segment (A) is 0.02 m0.02 \text{ m}.
    The time interval for each segment is 0.02 s0.02 \text{ s}.

  2. Calculate the initial velocity using the formula u=distancetimeu = \frac{\text{distance}}{\text{time}}.

    u=0.02 m0.02 su = \frac{0.02 \text{ m}}{0.02 \text{ s}} u=1.0 m/su = 1.0 \text{ m/s}

    The initial velocity is 1.0 m/s\boxed{1.0 \text{ m/s}}.

    b) ii) What is the final velocity, v, of the toy car?

  3. Identify the distance and time for the last segment.

    The distance of the last segment (B) is 0.06 m0.06 \text{ m}.
    The time interval for each segment is 0.02 s0.02 \text{ s}.

  4. Calculate the final velocity using the formula v=distancetimev = \frac{\text{distance}}{\text{time}}.

    v=0.06 m0.02 sv = \frac{0.06 \text{ m}}{0.02 \text{ s}} v=3.0 m/sv = 3.0 \text{ m/s}

    The final velocity is 3.0 m/s\boxed{3.0 \text{ m/s}}.

    c) Determine the acceleration of the toy car over the measured portion.

  5. Determine the total time elapsed for the change in velocity.

    There are 5 intervals shown on the tape. The time between the midpoint of the first interval and the midpoint of the fifth interval is (5−1)×time interval(5-1) \times \text{time interval}.

    t=4×0.02 st = 4 \times 0.02 \text{ s} t=0.08 st = 0.08 \text{ s}
  6. Calculate the acceleration using the formula a=v−uta = \frac{v - u}{t}.

    a=3.0 m/s−1.0 m/s0.08 sa = \frac{3.0 \text{ m/s} - 1.0 \text{ m/s}}{0.08 \text{ s}} a=2.0 m/s0.08 sa = \frac{2.0 \text{ m/s}}{0.08 \text{ s}} a=25 m/s2a = 25 \text{ m/s}^2

    The acceleration is 25 m/s2\boxed{25 \text{ m/s}^2}.

    6. Thermometer Calibration

    a) Give the name and values of the temperatures A and B in degree Celsius.

    • Temperature A: This represents the upper fixed point of the thermometer, which is the temperature of pure boiling water at standard atmospheric pressure. Its value is 100∘C\boxed{100^\circ \text{C}}.
    • Temperature B: This represents the lower fixed point of the thermometer, which is the temperature of pure melting ice at standard atmospheric pressure. Its value is $\boxed{0^\
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