Question

Question based on equation of motion​

Answer 1

Answer:

Explanation: 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!

Explanation: In a velocity time graph if we have to find out the distance or displacement then we have to find out the area under the curve!

Required solution:

a) Acceleration of the object

Between 0 to 2 seconds

Here, final velocity = 20 mps, initial velocity = 0 mps and time taken = 2 seconds (2-0)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{20-0}{2} \\ \\ :\implies \sf a \: = \dfrac{20}{2} \\ \\ :\implies \sf a \: = 10 \: mps^{-2} \\ \\ :\implies \sf Acceleration \: = 10 \: mps^{-2} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

Between 2 to 6 seconds

Here final velocity is 20 mps, initial velocity is 0 mps and time taken is 4 sec (6-2)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{20-0}{4} \\ \\ :\implies \sf a \: = \dfrac{20}{4} \\ \\ :\implies \sf a \: = 5 \: mps^{-2} \\ \\ :\implies \sf Acceleration \: = 5 \: mps^{-2} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

Between 6 to 8 seconds

Here, final velocity = 20 mps, initial velocity = 0 mps and time taken = 2 seconds (8-6)

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{20-0}{2} \\ \\ :\implies \sf a \: = \dfrac{20}{2} \\ \\ :\implies \sf a \: = 10 \: mps^{-2} \\ \\ :\implies \sf Acceleration \: = 10 \: mps^{-2} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

b) The total displacement by the object

Finding area of triangle first

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \: B \times H \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times (2-0) \times (20-0) \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times 2 \times 20 \\ \\ :\implies \sf Area \: of \: \triangle \: = 1 \times 1 \times 20 \\ \\ :\implies \sf Area \: of \: \triangle \: = 20 \: m^2$

Finding area of rectangle

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Area \: of \: rectangle \: = L \times B \\ \\ :\implies \sf Area \: of \: rectangle \: = (20-0) \times (6-2) \\ \\ :\implies \sf Area \: of \: rectangle \: = 20 \times 4 \\ \\ :\implies \sf Area \: of \: rectangle \: = 80 \: m^2$

Now calculating the area of triangle second

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \: B \times H \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times (8-6) \times (20-0) \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times 2 \times 20 \\ \\ :\implies \sf Area \: of \: \triangle \: = 1 \times 1 \times 20 \\ \\ :\implies \sf Area \: of \: \triangle \: = 20 \: m^2$

Now let's find total displacement.

$:\implies \sf Total \: displacement \: = Area(s) \\ \\ :\implies \sf Total \: displacement \: = 20 + 80 + 20 \\ \\ :\implies \sf Total \: displacement \: = 100 + 20 \\ \\ :\implies \sf Total \: displacement \: = 120 \: m \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$