Question

The velocity-time graph of a body is shown below. The displacement covered by body in 5 s is

A. 2m
B. 4m
C. 3m
D. 1m

Answer 1

Hint:If we find the value of the area in the velocity time graph we get the resultant value of displacement. On the other hand, the area count is defined as the distance covered by the body.

Formula used:
Area of triangle is given as
\[\dfrac{1}{2} \times b \times h\]
Where b is base and h is height.
Area of square=side\[ \times \]side

Complete step by step solution:
 In the velocity-time graph of a body as shown in the figure, the x-axis represents time in the second and the y-axis represents velocity in m/s.
Let from 0 s to 1 s area is \[{A_1}\], Area of square=side\[ \times \]side=\[1 \times 1\]
from 1 s to 2 s area is \[{A_2}\], Area of triangle=\[\dfrac{1}{2} \times b \times h\]
from 2 s to 3 s area is \[{A_3}\], Area of triangle=\[\dfrac{1}{2} \times b \times h\]
from 3 s to 4 s area is \[{A_4}\], Area of square=side\[ \times \]side=\[1 \times 1\]
from 4 s to 5 s area is \[{A_5}\], Area of square=side\[ \times \]side=\[ - (1 \times 1)\] , Here the negative sign shows time axis is the negative so velocity is in negative direction.

As we know the area enclosed in the velocity-time graph is equal to the displacement. Now the displacement (d) covered by the body in 5 s is calculated as,
Displacement,
\[d = {A_1} + {A_2} + {A_3} + {A_4} + {A_5}\]
\[\Rightarrow d = (1 \times 1) + \left( {\dfrac{1}{2} \times 1 \times 1} \right) + \left( {\dfrac{1}{2} \times 1 \times 1} \right) + (1 \times 1) - (1 \times 1)\]
$\therefore d= 2\,m$
Therefore, the displacement covered by the body in 5 s is 2 m.

Hence option A is the correct answer.

Note:A velocity-time graph can show the speed as well as direction an object can travel within a definite period of time. Velocity-time graphs are also known as speed-time graphs. The y-axis of a velocity-time graph represents the velocity of the object and the x-axis represents the time of the object.