Question

Please answer the questions above with process
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Answer 1

According to the given velocity time graph, we are asked to calculate or find the following,

a) Identify the type of motion the object has in the section AB and in the section BC.

Solution:

↪️ The type of motion the object has in the section AB and in the section BC are uniform motion and non-uniformly accelerated motion respectively.

→ AB performs uniform motion.

→ BC performs non-uniformly accelerated motion.

Information:

• Also in the section AB the velocity of the object is constant.

• And in the section BC the velocity of the body is decreasing, it starts retardation.

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b) Calculate the acceleration caused by the body.

Solution:

In section OA…

→ Here, final velocity is 60 m/s and initial velocity is 0 m/s and the the time taken is 15 seconds (because 15-0 = 15)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{60-0}{15} \\ \\ :\implies \sf a \: = \dfrac{60}{15} \\ \\ :\implies \sf a \: = \cancel{\dfrac{60}{15}} \\ \\ :\implies \sf Acceleration \: = 4 \: ms^{-2}$

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If talking about section AB then,

→ Here, final velocity is 60 m/s and initial velocity is 60 m/s and the the time taken is 30 seconds (because 45-15 = 30)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{60-60}{30} \\ \\ :\implies \sf a \: = \dfrac{0}{30} \\ \\ :\implies \sf a \: = 0 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 0 \: ms^{-2}$

Explanation: Surprised that acceleration is zero m/s sq. here? Actually section AB is not telling us about actual acceleration. As acceleration is defined as the change in velocity per time but in this section the final and initial velocity is same. And we also already know that is velocity is constant then acceleration is 0!

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For the part c) for this question kindly refer to attachment first.

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d) Calculate the deceleration caused by the body.

Solution:

→ Here, final velocity is 0 m/s and initial velocity is 60 m/s and the the time taken is 30 seconds (because 75 - 45 = 15)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-60}{30} \\ \\ :\implies \sf a \: = \dfrac{-60}{30} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-60}{30}} \\ \\ :\implies \sf a \: = -2 \\ \\ :\implies \sf Acceleration \: = -2 \: ms^{-2} \\ \\ :\implies \sf Deceleration \: = -2 \: ms^{-2}$

Explanation: Deceleration is the inverse of acceleration. And in the velocity time graph if the slope starts decreasing then it indicates deceleration.

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e) Find the distance travelled by the body in the entire 75 seconds.

Knowledge required:

• Area of triangle = ½ × B × H
• Area of rectangle = l × b

Where, B denotes base, H denotes height, l denotes length and b denotes breadth.

Explanation: We are able to see that the graph is made up of three shapes that are triangle, rectangle and again traingle. So we have to find out the area of these three shapes afterthat have to add them, to find our final result!

Solution:

~ Let's find out them individually.

Area of first triangle…

$:\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times Base \times Height \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times B \times H \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times 15 \times 60 \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times 900 \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{\cancel{{2}}} \times \cancel{900} \\ \\ :\implies \sf Area \: of \: \triangle \: = 450 \: m^2$

Let's find the area of rectangle…

$:\implies \sf Area \: of \: rectangle \: = Length \times Breadth \\ \\ :\implies \sf Area \: of \: rectangle \: = l \times b \\ \\ :\implies \sf Area \: of \: rectangle \: = 60 \times 30 \: (\because \: 45-15) \\ \\ :\implies \sf Area \: of \: rectangle \: = 1800 \: m^2$

Area of second triangle…

$:\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times Base \times Height \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times B \times H \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times 30 \: (\because \: 75-45) \times 60 \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times 1800 \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{\cancel{{2}}} \times \cancel{1800} \\ \\ :\implies \sf Area \: of \: \triangle \: = 900 \ m^2$

Now let's add them together to get distance travelled!

$:\implies \sf 450 + 1800 + 900 \\ \\ :\implies \sf 2250 + 900 \\ \\ :\implies \sf 3150 \: metres$