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Question 3 [5 marks] Find the total distance travelled, the distance travelled in the first 40 seconds, the acceleration when t = 25.

Mathematics

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Answers

3 parts
a

2200 m\text{2200 m}

b

1500 m\text{1500 m}

c

0 m/s2\text{0 m/s}^2

Step-by-step solution

9 steps

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

Part a· 4 steps

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 a trapezium.

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

    The base is 10 s10 \text{ s} and the height is 40 m/s40 \text{ m/s}.

    Area1=12×base×height=12×10 s×40 m/s=200 m\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \text{ s} \times 40 \text{ m/s} = 200 \text{ m}

  2. Calculate the area of the second section (rectangle from t=10t=10 to t=30t=30 s).

    The width is 30 s−10 s=20 s30 \text{ s} - 10 \text{ s} = 20 \text{ s} and the height is 40 m/s40 \text{ m/s}.

    Area2=width×height=20 s×40 m/s=800 m\text{Area}_2 = \text{width} \times \text{height} = 20 \text{ s} \times 40 \text{ m/s} = 800 \text{ m}

  3. Calculate the area of the third section (trapezium from t=30t=30 to t=50t=50 s).

    The parallel sides are 40 m/s40 \text{ m/s} (at t=30t=30) and 80 m/s80 \text{ m/s} (at t=50t=50). The height (time interval) is 50 s−30 s=20 s50 \text{ s} - 30 \text{ s} = 20 \text{ s}.

    Area3=12×(sum of parallel sides)×height=12×(40 m/s+80 m/s)×20 s=12×120 m/s×20 s=1200 m\text{Area}_3 = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (40 \text{ m/s} + 80 \text{ m/s}) \times 20 \text{ s} = \frac{1}{2} \times 120 \text{ m/s} \times 20 \text{ s} = 1200 \text{ m}

  4. Calculate the total distance.

    Total Distance=Area1+Area2+Area3=200 m+800 m+1200 m=2200 m\text{Total Distance} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3 = 200 \text{ m} + 800 \text{ m} + 1200 \text{ m} = 2200 \text{ m}

    The total distance travelled is 2200 m\boxed{\text{2200 m}}.

Part b

1500 m\text{1500 m}

The distance travelled in the first 40 seconds

This is the area under the graph from t=0t=0 to t=40t=40 s. This includes the first two sections and part of the third section.

  1. Determine the velocity at t=40t=40 s.

    The acceleration between t=30t=30 s and t=50t=50 s is constant.

    a=ΔvΔt=80 m/s−40 m/s50 s−30 s=40 m/s20 s=2 m/s2a = \frac{\Delta v}{\Delta t} = \frac{80 \text{ m/s} - 40 \text{ m/s}}{50 \text{ s} - 30 \text{ s}} = \frac{40 \text{ m/s}}{20 \text{ s}} = 2 \text{ m/s}^2

    Using v=u+at′v = u + at', where u=40 m/su=40 \text{ m/s} at t′=0t'=0 (which corresponds to t=30t=30 s), and t′=10t'=10 s (which corresponds to t=40t=40 s):

    v40=40 m/s+(2 m/s2×10 s)=40 m/s+20 m/s=60 m/sv_{40} = 40 \text{ m/s} + (2 \text{ m/s}^2 \times 10 \text{ s}) = 40 \text{ m/s} + 20 \text{ m/s} = 60 \text{ m/s}

  2. Calculate the area of the third part up to t=40t=40 s (trapezium from t=30t=30 to t=40t=40 s).

    The parallel sides are 40 m/s40 \text{ m/s} (at t=30t=30) and 60 m/s60 \text{ m/s} (at t=40t=40). The height (time interval) is 40 s−30 s=10 s40 \text{ s} - 30 \text{ s} = 10 \text{ s}.

    Area30−40=12×(40 m/s+60 m/s)×10 s=12×100 m/s×10 s=500 m\text{Area}_{30-40} = \frac{1}{2} \times (40 \text{ m/s} + 60 \text{ m/s}) \times 10 \text{ s} = \frac{1}{2} \times 100 \text{ m/s} \times 10 \text{ s} = 500 \text{ m}

  3. Calculate the total distance in the first 40 seconds.

    Distance0−40=Area1+Area2+Area30−40=200 m+800 m+500 m=1500 m\text{Distance}_{0-40} = \text{Area}_1 + \text{Area}_2 + \text{Area}_{30-40} = 200 \text{ m} + 800 \text{ m} + 500 \text{ m} = 1500 \text{ m}

    The distance travelled in the first 40 seconds is 1500 m\boxed{\text{1500 m}}.

Part c

0 m/s2\text{0 m/s}^2

The acceleration when t=25t = 25

Acceleration is the gradient (slope) of the velocity-time graph.

  1. Identify the segment of the graph at t=25t=25 s.

    At t=25t=25 s, the graph is a horizontal line segment between t=10t=10 s and t=30t=30 s.

  2. Calculate the gradient of this segment.

    In this segment, the velocity is constant at 40 m/s40 \text{ m/s}. A horizontal line has a gradient of zero.

    a=ΔvΔt=40 m/s−40 m/s30 s−10 s=0 m/s20 s=0 m/s2a = \frac{\Delta v}{\Delta t} = \frac{40 \text{ m/s} - 40 \text{ m/s}}{30 \text{ s} - 10 \text{ s}} = \frac{0 \text{ m/s}}{20 \text{ s}} = 0 \text{ m/s}^2

    The acceleration when t=25t=25 s is 0 m/s2\boxed{\text{0 m/s}^2}.

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