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The graph below illustrates the speed of a car in a period of 16 seconds as it drives along a gravelled road. (a) Find (i) the speed of the car when t = 4, (ii) the acceleration when t = 12. (b) Calculate the total distance travelled during the 16 seconds.

Physics

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Answers

3 parts
a(i)

15 m/s\text{15 m/s}

a(ii)

0 m/s2\text{0 m/s}^2

b

200 m\text{200 m}

Step-by-step solution

8 steps

Part a(i)· 2 steps

The speed of the car when t=4t=4 seconds.

  1. Identify the coordinates on the graph.

    The graph is a straight line from (0,20)(0, 20) to (8,10)(8, 10). To find the speed at t=4t=4, we can find the equation of this line or use similar triangles/interpolation.
    The slope (acceleration) for 0≤t≤80 \le t \le 8 is:

    a=10−208−0=−108=−1.25 m/s2a = \frac{10 - 20}{8 - 0} = \frac{-10}{8} = -1.25 \text{ m/s}^2

    The equation of the line is v=−1.25t+20v = -1.25t + 20.

  2. Substitute t=4t=4 into the equation.

    v=−1.25(4)+20v = -1.25(4) + 20 v=−5+20v = -5 + 20 v=15 m/sv = 15 \text{ m/s}

    The speed of the car when t=4t=4 is 15 m/s\boxed{\text{15 m/s}}.

Part a(ii)

0 m/s2\text{0 m/s}^2

The acceleration when t=12t=12 seconds.

  1. Identify the section of the graph for t=12t=12.

    For the time interval 8≤t≤168 \le t \le 16, the graph is a horizontal line at a constant speed of 10 m/s10 \text{ m/s}.

  2. Calculate the acceleration.

    Acceleration is the gradient of the speed-time graph. Since the speed is constant, the gradient is zero.

    a=ΔvΔt=10−1016−8=08=0 m/s2a = \frac{\Delta v}{\Delta t} = \frac{10 - 10}{16 - 8} = \frac{0}{8} = 0 \text{ m/s}^2

    The acceleration when t=12t=12 is 0 m/s2\boxed{\text{0 m/s}^2}.

Part b

200 m\text{200 m}

Calculate the total distance travelled during the 16 seconds.

  1. The total distance travelled is the area under the speed-time graph.

    The area can be divided into two parts: a trapezium from t=0t=0 to t=8t=8 and a rectangle from t=8t=8 to t=16t=16.

  2. Calculate the area of the trapezium (0≤t≤80 \le t \le 8).

    The parallel sides are v1=20 m/sv_1 = 20 \text{ m/s} (at t=0t=0) and v2=10 m/sv_2 = 10 \text{ m/s} (at t=8t=8). The height is Δt=8−0=8 s\Delta t = 8 - 0 = 8 \text{ s}.

    Area1=12(v1+v2)Δt=12(20+10)(8)\text{Area}_1 = \frac{1}{2}(v_1 + v_2)\Delta t = \frac{1}{2}(20 + 10)(8) Area1=12(30)(8)=15×8=120 m\text{Area}_1 = \frac{1}{2}(30)(8) = 15 \times 8 = 120 \text{ m}

  3. Calculate the area of the rectangle (8≤t≤168 \le t \le 16).

    The width is Δt=16−8=8 s\Delta t = 16 - 8 = 8 \text{ s}. The height (speed) is 10 m/s10 \text{ m/s}.

    Area2=width×height=8×10=80 m\text{Area}_2 = \text{width} \times \text{height} = 8 \times 10 = 80 \text{ m}

  4. Calculate the total distance.

    Total Distance=Area1+Area2=120 m+80 m=200 m\text{Total Distance} = \text{Area}_1 + \text{Area}_2 = 120 \text{ m} + 80 \text{ m} = 200 \text{ m}

    The total distance travelled during the 16 seconds is 200 m\boxed{\text{200 m}}.

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