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 :
![](https://imagex.cdn.examgoal.net/jgy2xukfl4ld7t/6a32aa8d-71f3-4cec-a806-e3c8ac9ccf1b/7f5b6190-00c3-11eb-8293-3b6cc591631c/file-jgy2xukfl4ld7u-320w.png)

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: