Question

the v-t graph of an object is shown below​

Answer 1

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!

→ 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) What type of motion is represented by OA, AB and BC!?

Answer:

In section OA : This section shows the uniform acceleration that is “uniform accelerated motion.”

In section AB : This section shows the “uniform motion” , constant velocity and zero acceleration.

In section BC : This section shows the “uniform retardation”.

(b) Find positive and negative acceleration of the body.

Finding position acceleration :

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{20-0}{10} \\ \\ :\implies \sf a \: = \dfrac{20}{10} \\ \\ :\implies \sf a \: = 2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2 \: ms^{-2}$

Finding negative acceleration :

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-20}{(40-30)} \\ \\ :\implies \sf a \: = \dfrac{-20}{10} \\ \\ :\implies \sf a \: = -2 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -2 \: ms^{-2}$

(c) Find distance travelled by the body from A to B.

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: of \: rectangle \\ \\ :\implies \sf Distance \: = Length \times Breadth \\ \\ :\implies \sf Distance \: = 20 \times (30-10) \\ \\ :\implies \sf Distance \: = 20 \times 20 \\ \\ :\implies \sf Distance \: = 400 \: m$