Question

In the given velocity-time graph, interpret the nature of the object’s motion throughout the journey.

Calculate the acceleration and displacement between time interval 0 to 4s.

Answer 1

From the given velocity time graph we are asked to calculate the acceleration and displacement between time interval 0 to 4 seconds.

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:

~ Firstly let us calculate the acceleration!

Acceleration from section O to A

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{6-0}{4-0} \\ \\ :\implies \sf a \: = \dfrac{6}{4} \\ \\ :\implies \sf a \: = \cancel{\dfrac{6}{4}} \: (Cancelling) \\ \\ :\implies \sf a \: = \dfrac{3}{2} \\ \\ :\implies \sf a \: = 1.5 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 1.5 \: ms^{-2} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

Now from section A to B

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{6-6}{10-4} \\ \\ :\implies \sf a \: = \dfrac{0}{6} \\ \\ :\implies \sf a \: = 0 \: ms^{-1} \\ \\ :\implies \sf Acceleration \: = 0 \: ms^{-2} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

There is zero acceleration here because as I say that if the slope is parallel to time axis then there is no acceleration taking place.

Now from section B to C

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-6}{16-10} \\ \\ :\implies \sf a \: = \dfrac{-6}{6} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-6}{6}} \: (Cancelling) \\ \\ :\implies \sf a \: = -1 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -1 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -1 \: ms^{-2} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Now let's calculate the displacement at the time interval of 0 to 4 seconds

$:\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf s \: = Area \: of \: \triangle \: AOD \\ \\ :\implies \sf s \: = \dfrac{1}{2} \: Base \times Height \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times (4-0) \times (6-0) \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 4 \times 6 \\ \\ :\implies \sf s \: = \dfrac{1}{2} \times 24 \\ \\ :\implies \sf s \: = \dfrac{1}{\cancel{{2}}} \times \cancel{24} \\ \\ :\implies \sf s \: = 12 \: m \\ \\ :\implies \sf Displacement \: = 12 \: m\\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$