Question

pls physics genius answer this​

Answer 1

Answer:

Question 30.

• According to figure 8.38 we are asked to find out the velocity in during the last 10 seconds!

»»» Last 10 seconds are between the time interval of 35 to 45.

We know that,

• ${\small{\underline{\boxed{\pmb{\sf{v \: = \dfrac{s}{t}}}}}}}$

Where, v denotes velocity, s denotes displacement or distance and t denotes time taken.

$:\implies \sf v \: = \dfrac{s}{t} \\ \\ :\implies \sf Velocity \: = \dfrac{Distance}{Time} \\ \\ :\implies \sf Velocity \: = \dfrac{70-40}{45-35} \\ \\ :\implies \sf Velocity \: = \dfrac{30}{10} \\ \\ :\implies \sf Velocity \: = 3 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

• Therefore, option (c) is right.

Question 31.

• We are asked to find out the distance travelled in 45 seconds according to above question!

Firstly let us find out the area of triangle!

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: \triangle \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times B \times H \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 20 \times 40 \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 800 \\ \\ :\implies \sf Distance \: = 400 \: m$

Finding area of rectangle!

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: rectangle \\ \\ :\implies \sf Distance \: = L \times B \\ \\ :\implies \sf Distance \: = 15 \times 40 \\ \\ :\implies \sf Distance \: = 600 \: m$

Now area of trapezium!

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: trapezium \\ \\ :\implies \sf Distance \: = \dfrac{a+b}{2} \times h \\ \\ :\implies \sf Distance \: = \dfrac{40+70}{2} \times 10 \\ \\ :\implies \sf Distance \: = \dfrac{110}{2} \times 10 \\ \\ :\implies \sf Distance \: = 55 \times 10 \\ \\ :\implies \sf Distance \: = 550 \: m$

Now finding total distance!

$:\implies \sf 400 + 600 + 550 \\ \\ :\implies \sf 1000 + 550 \\ \\ :\implies \sf 1550 \: m \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

• Therefore, option (d) is right.

Question 32.

• According to figure 8.39 we are asked to determine the time in which the marble is in motion, it travel A first then the horizontal way afterthat the section B!

According to the marble is in motion for 100 seconds.

Question 33.

• According to question 32, what is the maximum speed of the marble?

Answer: Maximum speed of marble is 20 metre per second.

Question 34.

• Total distance travelled is, in question 32!?

Finding area of triangle!

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: \triangle \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times B \times H \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 20 \times 20 \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 400 \\ \\ :\implies \sf Distance \: = 200 \: m$

Finding area of rectangle!

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: rectangle \\ \\ :\implies \sf Distance \: = L \times B \\ \\ :\implies \sf Distance \: = 30 \times 20 \\ \\ :\implies \sf Distance \: = 600 \: m$

Finding area of triangle second!

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: \triangle \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times B \times H \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 50 \times 20 \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 1000 \\ \\ :\implies \sf Distance \: = 500 \: m$

Finding total distance!

$:\implies \sf 200 + 600 + 500 \\ \\ :\implies \sf 200 + 1100 \\ \\ :\implies \sf 1300 \: m \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

• Therefore, option (d) is correct.