Question

1. Use below figure to find :
(i) total distance covered in 1.5 s.
retardation from 0.5 s to 1.5

Answer 1

Solution 1st

From the given velocity time graph we are asked to calculate

1) Total distance covered in 1.5 seconds

Firstly let us calculate the area of rectangle!

$:\implies \sf Area \: of \: rectangle \: = Length \times Breadth \\ \\ :\implies \sf Area \: of \: rectangle \: = (0.5 - 0) \times (10 - 0) \\ \\ :\implies \sf Area \: of \: rectangle \: = 0.5 \times 10 \\ \\ :\implies \sf Area \: of \: rectangle \: = 5 \: unit \: sq.$

Now let us find out the area of given traingle!

$:\implies \sf Area \: of \: triangle \: = \dfrac{1}{2} \: Base \times Height \\ \\ :\implies \sf Area \: of \: triangle \: = \dfrac{1}{2} \times (1.5-0.5) \times (10-0) \\ \\ :\implies \sf Area \: of \: triangle \: = \dfrac{1}{2} \times 1 \times 10 \\ \\ :\implies \sf Area \: of \: triangle \: = \dfrac{1}{2} \times 10 \\ \\ :\implies \sf Area \: of \: triangle \: = 5 \: unit \: sq.$

Now let us calculate total distance

$:\implies \sf 5 + 5 \\ \\ :\implies \sf Total \: distance \: = 10 \: m$

2) Now let us calculate retardation from 0.5 seconds to 1.5 seconds!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-10}{1.5-0.5} \\ \\ :\implies \sf a \: = \dfrac{-10}{1} \\ \\ :\implies \sf a \: = -10 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -10 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -10 \: ms^{-2}$

Solution 2nd

Second equation of motion:

Firstly let, the object is travelling a distance in time under uniform acceleration. Now according to the graph we are able to see that inside the graph the obtained area enclose within OABC(trapezium) under velocity time graph AB. Therefore, distance travelled by an object can be given by

$\tt \Rightarrow Distance \: = Area \: enclosed \\ \\ \tt \Rightarrow s \: = Area \: enclosed \\ \\ \tt \Rightarrow s \: = OABC \\ \\ \tt \Rightarrow s \: = Area \: of \: rectangle \: + Area \: of \: \triangle \\ \\ \tt \Rightarrow s \: = Length \times Breadth + \dfrac{1}{2} \times Base \times Height \\ \\ \tt \Rightarrow s \: = AO \times AD + \dfrac{1}{2} \times AD \times BD \\ \\ \tt \Rightarrow s \: = u \times t + \dfrac{1}{2} \times t \times BD \\ \\ \tt \Rightarrow s \: = u \times t + \dfrac{1}{2} \times t \times at \\ \\ \tt \Rightarrow s \: = ut + \dfrac{1}{2} \times at^2 \\ \\ {\pmb{\sf{Henceforth, \: derived!}}}$

How the value of BD came?

Firstly we can write BC as BD + DC. Now as BD have the velocity position and DC have the time position. Henceforth, we already know that

$\tt \Rightarrow Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{v-u}{t} \\ \\ \tt \Rightarrow a \: = \dfrac{BD}{t} \\ \\ \tt \Rightarrow at \: = BD$