Question 68: Motion in 1D

Question

The velocity (v) and time (t) graph of a body in
a straight line motion is shown in the figure.
The point S is at 4.333 seconds. The total
distance covered by the body in 6 s is :

Accepted Answer

4.333 sec = $${{13} \over 3}$$ sec
Distance = area under the v-t graph
= Area of Parallelogram + Area of triangle
= $${1 \over 2}\left( 4 ight)\left( {{{13} \over 3} + 1} ight)$$ + $${1 \over 2}\left( {6 - {{13} \over 3}} ight) 	imes 2$$
= $${{37} \over 3}$$ m

Answer

12 m

Answer

11 m

Answer

$${{49} \over 4}$$ m

Answer

$${{37} \over 3}$$ m

Motion in 1D

The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is :

Choose the correct answer: