Question

please help me in this graph.And please give a step by step explanation.
1.What is the acceleration of particle in portion ab and bc.
2.What is total distance of the particle.
3.What is average velocity of the particle.​

Answer 1

According to the given velocity time graph we are asked to solve some questions.

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:

a) Acceleration of the given partical in portion ab and bc

In portion ab

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

Explanation: If the slope of Velocity Time graph is parallel to time axis then the acceleration is always zero as there is no change in velocity as acceleration is the term that tell us about the change in velocity with respect to time.

In portion bc

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{80-40}{60-40} \\ \\ :\implies \sf a \: = \dfrac{40}{20} \\ \\ :\implies \sf a \: = 2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2 \: ms^{-2}$

b) Total distance of the particle

Firstly finding area of 1st triangle

$:\implies \sf Area \: = \dfrac{1}{2} \: Base \times Height \\ \\ :\implies \sf Area \: = \dfrac{1}{2} \times (20-0) \times (40-0) \\ \\ :\implies \sf Area \: = \dfrac{1}{2} \times 20 \times 40 \\ \\ :\implies \sf Area \: = \dfrac{1}{2} \times 800 \\ \\ :\implies \sf Area \: = \dfrac{1}{\cancel{{2}}} \times \cancel{800} \\ \\ :\implies \sf Area \: = 400 \: unit =\: sq.$

Finding area of rectangle 1st

$:\implies \sf Area \: = Length \times Breadth \\ \\ :\implies \sf Area \: = (40-20) \times (40-0) \\ \\ :\implies \sf Area \: = 20 \times 40 \\ \\ :\implies \sf Area \: = 800 \: unit \: sq.$

Finding area of trapezium 1st

$:\implies \sf Area \: = \dfrac{Sum \: of \: parallel \: sides}{2} \times Height \\ \\ :\implies \sf Area \: = \dfrac{80+40}{2} \times (60-40) \\ \\ :\implies \sf Area \: = \dfrac{80+40}{2} \times 20 \\ \\ :\implies \sf Area \: = \dfrac{80+40}{2} \times 20 \\ \\ :\implies \sf Area \: = \dfrac{120}{2} \times 20 \\ \\ :\implies \sf Area \: = 60 \times 20 \\ \\ :\implies \sf Area \: = 1200 \: unit \: sq.$

Finding area of 1st triangle

$:\implies \sf Area \: = \dfrac{1}{2} \: Base \times Height \\ \\ :\implies \sf Area \: = \dfrac{1}{2} \times (100-60) \times (80-0) \\ \\ :\implies \sf Area \: = \dfrac{1}{2} \times 40 \times 80 \\ \\ :\implies \sf Area \: = \dfrac{1}{2} \times 3200 \\ \\ :\implies \sf Area \: = 1600 \: unit \: sq.$

Now let us calculate distance

$:\implies \sf 400 + 800 + 1200 + 1600 \\ \\ :\implies \sf Distance \: = 4000 \: m$

c) Average velocity of the particle

Explanation: We know that average velocity = Total distance/Total time Therefore,

• Total distance = 4000 m
• Total time = 100 seconds

Henceforth,

→ Average velocity = Total distance/Total time

→ Average velocity = 4000/100

→ Average velocity = 40 m/s