Question

PLs find it. the attachment is given below

Answer 1

From the given graph, we are asked to calculate the acceleration:

• During the first two seconds
• Between 2nd and 10th second
• During the last two seconds

KnowlEdge required:

As we already know that the given graph is velocity time graph and we get the accelaration from velocity time graph by it's slope. Acceleration's formula:

• ${\small{\underline{\boxed{\sf{a \: = \dfrac{v-u}{t}}}}}}$

Where, a denotes acceleration, v denotes final velocity, u denotes initial velocity and t denotes the time taken.

RequirEd solution:

1) During the first two seconds

↪️ Here, initial velocity is 0 m/s, final velocity is 4.6 m/s and time taken is 2 seconds (∵ 2-0 = 2)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time \: taken} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{4.6-0}{2} \\ \\ :\implies \sf a \: = \dfrac{4.6}{2} \\ \\ :\implies \sf a \: = 2.3 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2.3 \: ms^{-2}$

∴ The acceleration during the first two seconds is 2.3 m/s²

2) Between 2nd and 10th second

• 1st method...

↪️ Here, initial velocity is 4.6 m/s, final velocity is 4.6 m/s and time taken is 8 seconds (∵ 10-2 = 8)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time \: taken} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{4.6-4.6}{8} \\ \\ :\implies \sf a \: = \dfrac{0}{8} \\ \\ :\implies \sf a \: = 0 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 0 \: ms^{-2}$

∴ The acceleration between 2nd and 10th second is 0 m/s²

• 2nd method...

The accelaration between 2nd and 10th second is 0 m/s² because the body isn't changing its velocity. And we already know that constant velocity = zero acceleration.

3) During the last two seconds

↪️ Here, initial velocity is 4.6 m/s, final velocity is 0 m/s and time taken is 2 seconds (∵ 12-10 = 2)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time \: taken} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-4.6}{2} \\ \\ :\implies \sf a \: = \dfrac{-4.6}{2} \\ \\ :\implies \sf a \: = -2.3 ms^{-2} \\ \\ :\implies \sf Acceleration \: = -2.3 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -2.3 \: ms^{-2}$

∴ The acceleration during the last two seconds is -2.3 m/s²

• It came in negative because the slope is going downwards