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A car's motion is represented by a speed-time graph. Analyze the motion at points R, S, and T, find the speed at 4.0 s, and calculate the distance traveled between 16.0 s and 20.0 s.

Physics

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2 parts
b

18 m/s\text{18 m/s}

c

40 m\text{40 m}

Step-by-step solution

5 steps

Part a

Describe the motion of the graph at points

  • (i) R: The car is accelerating, but its acceleration is decreasing. The speed is increasing, but the rate at which it increases is slowing down.
  • (ii) S: The car is moving at a constant speed. Its acceleration is zero.
  • (iii) T: The car is decelerating (slowing down) at a constant rate. Its acceleration is constant and negative.

Part b

18 m/s\text{18 m/s}

Determine the speed of the car at time = 4.0 s.

  1. Locate t=4.0t = 4.0 s on the time axis.

  2. Read the corresponding speed value from the graph.

    At t=4.0t = 4.0 s, the speed is 18 m/s.

    The speed of the car at t=4.0t = 4.0 s is 18 m/s\boxed{\text{18 m/s}}.

Part c

40 m\text{40 m}

Determine the distance moved by the car from time = 16.0 s to time = 20.0 s.

The distance moved is the area under the speed-time graph. For the interval from t=16.0t = 16.0 s to t=20.0t = 20.0 s, the shape is a triangle.

  1. Identify the coordinates at the start and end of the interval.

    At t1=16.0t_1 = 16.0 s, speed v1=20v_1 = 20 m/s.
    At t2=20.0t_2 = 20.0 s, speed v2=0v_2 = 0 m/s.

  2. Calculate the base and height of the triangle.

    Base =Δt=t2−t1=20.0 s−16.0 s=4.0 s= \Delta t = t_2 - t_1 = 20.0 \text{ s} - 16.0 \text{ s} = 4.0 \text{ s}.
    Height =v1=20= v_1 = 20 m/s (since v2=0v_2 = 0).

  3. Calculate the area of the triangle.

    Distance=12×base×height\text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} Distance=12×(4.0 s)×(20 m/s)\text{Distance} = \frac{1}{2} \times (4.0 \text{ s}) \times (20 \text{ m/s}) Distance=2 s×20 m/s\text{Distance} = 2 \text{ s} \times 20 \text{ m/s} Distance=40 m\text{Distance} = 40 \text{ m}

    The distance moved by the car from t=16.0t = 16.0 s to t=20.0t = 20.0 s is 40 m\boxed{\text{40 m}}.

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