Question

# The following plot shows the velocity-time relation of a moving body. Then, pick the CORRECT statement(s) from the following.

A
The body decelerated at a rate of 5 m s2 in the last two seconds of its motion.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
The body is displaced by 62.5 cm.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
The body accelerated at a rate of 5 m s2 in the first four seconds of its motion.
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
The body has uniformly accelerated motion from a total of 0 to 4 s.
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is B The body is displaced by 62.5 cm.The slope of velocity-time graph of a body gives the acceleration of the body. Acceleration from time (t) = 3 s to t=4 s is 10−54−3=5 m s−2 ⟹ The body is accelerating from t=3 s to t=4 s, i.e. for one second. And, acceleration from time (t) = 7 s to t=9 s is 0−109−7=−5 m s−2 An increasing slope indicates that the body is accelerating, while a decreasing one indicates deceleration. Also, from t=0 s to t=3 s, the slope is zero and from t=3 s to t=4 s, the slope is non-zero and constant. Hence, the body has non-uniform acceleration for the interval t=0 s to t=4 s. The area under the velocity-time graph gives the total (net) displacement of the body: The sum of the areas of the sections A to E gives the net displacement. ⟹ Net displacement = 15 cm+5 cm+2.5 cm+30 cm+10 cm=62.5 cm

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Acceleration
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program