Question

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Answer 1

According to the given velocity time graph we are asked to determine the following, let's solve this question!

Knowledge required:

• In a velocity time graph, the slope of graph tell us about the acceleration. If the slope is high then the acceleration is positive, if the solve is low then the acceleration is negative and if the slope is parallel to time axis then there is no acceleration taking place!

• In the velocity time graph the distance or displacement can be founded by the area under the curve!

Required solution:

Required solution: a) The type of acceleration represented by the following:

• Section AB
• Section BD
• Section DE

Solution: The type of acceleration represented by the following:

• Section AB = Positive acceleration

• Section BD = Zero acceleration

• Section DE = Negative acceleration

b) Acceleration in the first two hours, in the next two hours and in the last two hours.

Acceleration in first two hours

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{40-0}{2-0} \\ \\ :\implies \sf a \: = \dfrac{40}{2} \\ \\ :\implies \sf a \: = \cancel{\dfrac{40}{2}} \: (Cancelling) \\ \\ :\implies \sf a \: = 20 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 20 \: ms^{-2}$

Now acceleration in next two hours!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{40-40}{4-2} \\ \\ :\implies \sf a \: = \dfrac{0}{2} \\ \\ :\implies \sf a \: = 0 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 0 \: ms^{-2}$

Acceleration in last two hours

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-40}{8-6} \\ \\ :\implies \sf a \: = \dfrac{-40}{2} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-40}{2}} \: (Cancelling) \\ \\ :\implies \sf a \: = -20 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -20 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -20 \: ms^{-2}$

c) Total distance travelled by the car

~ Let us calculate the area of each figure individually.

Area of triangle ABF

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: of \: \triangle ABF \\ \\ :\implies \sf s \: = \dfrac{1}{2} \: Base \times Height \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times (2-0) \times (40-0) \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 2 \times 40 \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 80 \\ \\ :\implies \sf s \: = \dfrac{1}{\cancel{{2}}} \times \cancel{80} \: (Cancelling) \\ \\ :\implies \sf s \: = 40 \: unit \: sq.$

Area of rectangle BCFO

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: of \: rectangle \\ \\ :\implies \sf s \: = Length \times Breadth \\ \\ :\implies \sf s \: = (40-0) \times (4-2) \\ \\ :\implies \sf s \: = 40 \times 2 \\ \\ :\implies \sf s \: = 80 \: unit \: sq.$

Area of rectangle CDOG

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: of \: rectangle \\ \\ :\implies \sf s \: = Length \times Breadth \\ \\ :\implies \sf s \: = (40-0) \times (6-4) \\ \\ :\implies \sf s \: = 40 \times 2 \\ \\ :\implies \sf s \: = 80 \: unit \: sq.$

Area of triangle DEG

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: of \: \triangle DEG \\ \\ :\implies \sf s \: = \dfrac{1}{2} \: Base \times Height \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times (8-6) \times (40-0) \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 2 \times 40 \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 80 \\ \\ :\implies \sf s \: = \dfrac{1}{\cancel{{2}}} \times \cancel{80} \: (Cancelling) \\ \\ :\implies \sf s \: = 40 \: unit \: sq.$

Now let us calculate total distance by add them together

$:\implies \sf 40 + 80 + 80 + 40 \\ \\ :\implies \sf 80 + 80 + 80 \\ \\ :\implies \sf 160 + 80 \\ \\ :\implies \sf Total \: distance \: = 240 \: kilometres$

d) Average speed of the car

$:\implies \sf Average \: speed \: = \dfrac{Total \: distance}{Total \: time} \\ \\ :\implies \sf Average \: speed \: = \dfrac{240}{8} \\ \\ :\implies \sf Average \: speed \: = 30 \: kmh^{-1}$