Question

it's very important that you find all the answers that includes a through e and show your work, thank you​

Answer 1

Answer:

Explanation:

▶️ To find acceleration in the velocity-time graph we have to use the slope of the graph. And then further we have to use the formula. 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!

▶️ To calculate the distance or displacement from a velocity-time graph we have to calculate the area under the curve!

Required solution:

a) Acceleration

Acceleration in section OA!

Here, initial velocity is 0 m/s, final velocity is 30 m/s and time taken is 8 seconds (because 8-0)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{30-0}{8} \\ \\ :\implies \sf a \: = \dfrac{30}{8} \\ \\ :\implies \sf a \: = \dfrac{15}{4} \\ \\ :\implies \sf a \: = 3.75 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 3.75 \: ms^{-2}$

b) Retardation

Retardation in section AB!

Here, final velocity is 0 m/s , initial velocity is 30 m/s and time taken is 8 (because 16-8)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-30}{8} \\ \\ :\implies \sf a \: = \dfrac{-30}{8} \\ \\ :\implies \sf a \: = \dfrac{-15}{4} \\ \\ :\implies \sf a \: = -3.75 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -3.75 \: ms^{-2}$

c) Distance covered in first 8 seconds

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: \triangle \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times B \times H \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times (8-0) \times (30-0) \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 8 \times 30 \\ \\ :\implies \sf Distance \: = 1 \times 4 \times 30 \\ \\ :\implies \sf Distance \: = 120 \: m$

d) Total distance travelled by the body

Area of first triangle AOC!

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: \triangle \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times B \times H \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times (8-0) \times (30-0) \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 8 \times 30 \\ \\ :\implies \sf Distance \: = 1 \times 4 \times 30 \\ \\ :\implies \sf Distance \: = 120 \: m$

• Here area = 120 m sq.

Area of second triangle ACB!

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: \triangle \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times B \times H \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times (16-8) \times (30-0) \\ \\ :\implies \sf Distance \: = \dfrac{1}{2} \times 8 \times 30 \\ \\ :\implies \sf Distance \: = 1 \times 4 \times 30 \\ \\ :\implies \sf Distance \: = 120 \: m$

• Here area = 120 m sq.

Therefore, total distance!

$:\implies \sf Total \: distance \: = Area \: 1 \: + Area \: 2 \\ \\ :\implies \sf Total \: distance \: = 120 + 120 \\ \\ :\implies \sf Total \: distance \: = 240 \: m$

e) Average velocity in the region ACB.

Explanation: We know that average velocity = Total displacement/Total time and here total displacement is 240 m and total time is 16 seconds.

$:\implies \sf a_{av} \: = \dfrac{Total \: displacement}{Total \: time} \\ \\ :\implies \sf a_{av} \: = \dfrac{240}{16} \\ \\ :\implies \sf a_{av} \: = 15 \: ms^{-1}$