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Question 5 [3 marks] Find: (a) the acceleration when t = 4, (b) the total distance travelled, (c) the average speed for the whole journey.

Mathematics

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Answers

3 parts
a

1.5 m/s2\text{1.5 m/s}^2

b

675 m\text{675 m}

c

11.25 m/s\text{11.25 m/s}

Step-by-step solution

8 steps

Here's how to solve the problem using the provided velocity-time graph.

Part a· 2 steps

The acceleration when t=4t = 4

Acceleration is the gradient of the velocity-time graph. At t=4t=4 s, the graph is in the first segment, which is a straight line from (0 s,0 m/s)(0 \text{ s}, 0 \text{ m/s}) to (10 s,15 m/s)(10 \text{ s}, 15 \text{ m/s}).

  1. Identify the coordinates of the segment.

    The initial point is (t1,v1)=(0 s,0 m/s)(t_1, v_1) = (0 \text{ s}, 0 \text{ m/s}) and the final point is (t2,v2)=(10 s,15 m/s)(t_2, v_2) = (10 \text{ s}, 15 \text{ m/s}).

  2. Calculate the acceleration using the gradient formula.

    a=ΔvΔt=v2−v1t2−t1=15 m/s−0 m/s10 s−0 s=15 m/s10 s=1.5 m/s2a = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1} = \frac{15 \text{ m/s} - 0 \text{ m/s}}{10 \text{ s} - 0 \text{ s}} = \frac{15 \text{ m/s}}{10 \text{ s}} = 1.5 \text{ m/s}^2

    The acceleration when t=4t=4 s is 1.5 m/s2\boxed{\text{1.5 m/s}^2}.

Part b

675 m\text{675 m}

The total distance travelled

The total distance travelled is the area under the velocity-time graph. We can divide the graph into three sections: a triangle, a rectangle, and another triangle.

  1. Calculate the area of the first section (triangle from t=0t=0 to t=10t=10 s).

    Area1=12×base×height=12×10 s×15 m/s=75 m\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \text{ s} \times 15 \text{ m/s} = 75 \text{ m}

  2. Calculate the area of the second section (rectangle from t=10t=10 to t=40t=40 s).

    Area2=width×height=(40 s−10 s)×15 m/s=30 s×15 m/s=450 m\text{Area}_2 = \text{width} \times \text{height} = (40 \text{ s} - 10 \text{ s}) \times 15 \text{ m/s} = 30 \text{ s} \times 15 \text{ m/s} = 450 \text{ m}

  3. Calculate the area of the third section (triangle from t=40t=40 to t=60t=60 s).

    Area3=12×base×height=12×(60 s−40 s)×15 m/s=12×20 s×15 m/s=150 m\text{Area}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (60 \text{ s} - 40 \text{ s}) \times 15 \text{ m/s} = \frac{1}{2} \times 20 \text{ s} \times 15 \text{ m/s} = 150 \text{ m}

  4. Calculate the total distance by summing the areas.

    Total Distance=Area1+Area2+Area3=75 m+450 m+150 m=675 m\text{Total Distance} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = 75 \text{ m} + 450 \text{ m} + 150 \text{ m} = 675 \text{ m}

    The total distance travelled is 675 m\boxed{\text{675 m}}.

Part c

11.25 m/s\text{11.25 m/s}

The average speed for the whole journey

Average speed is the total distance travelled divided by the total time taken.

  1. Identify the total distance and total time.

    Total distance = 675 m675 \text{ m} (from part b)
    Total time = 60 s60 \text{ s} (from the graph)

  2. Calculate the average speed.

    Average Speed=Total DistanceTotal Time=675 m60 s=11.25 m/s\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{675 \text{ m}}{60 \text{ s}} = 11.25 \text{ m/s}

    The average speed for the whole journey is 11.25 m/s\boxed{\text{11.25 m/s}}.

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