Answer:
Explanation:
i) during first two seconds
v/t =a so slope graph is acceleration
slope of graph =tan theta
y/x= 4.6 /2= 2.3 m/s square
iii) similarly last two seconds
but it deaccelerates so it will be in negative sign
tan theta= -4.6/2=-2.3 m/s square
ii) as slope of graph constant , tan theta =dy /dx
from differentiation
acceleration =0
----------------------------------------------------------------------------------
For the first two seconds --
Acceleration = Change in velocity/Time
Acceleration= (IMPORTANT)
By observing the graph--
v=4.6 m/s
u=0 m/s
t= 2 secs
For the last two seconds--
v= 0 m/s
u= 4.6 m/s
a=
Between 2nd and 10th seconds--
Since the line is a straight line parallel to the time axis
The velocity is constant (velocity does not increases or decreases)
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Answers & Comments
Answer:
Explanation:
i) during first two seconds
v/t =a so slope graph is acceleration
slope of graph =tan theta
y/x= 4.6 /2= 2.3 m/s square
iii) similarly last two seconds
but it deaccelerates so it will be in negative sign
tan theta= -4.6/2=-2.3 m/s square
ii) as slope of graph constant , tan theta =dy /dx
from differentiation
acceleration =0
Hello!
Nice Question
----------------------------------------------------------------------------------
For the first two seconds --
Acceleration = Change in velocity/Time
Acceleration=
(IMPORTANT)
By observing the graph--
v=4.6 m/s
u=0 m/s
t= 2 secs
Hence, a=
= 2.3 m/s^2
For the last two seconds--
v= 0 m/s
u= 4.6 m/s
t= 2 secs
a=![\frac{(v-u)}{t} \frac{(v-u)}{t}](https://tex.z-dn.net/?f=%5Cfrac%7B%28v-u%29%7D%7Bt%7D)
a=![\frac{0-4.6}{2} \frac{0-4.6}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B0-4.6%7D%7B2%7D)
a= -2.3 m/s^2 (Negative sign indicates deceleration)
Between 2nd and 10th seconds--
Since the line is a straight line parallel to the time axis
The velocity is constant (velocity does not increases or decreases)
Hence, Acceleration=0 m/s^2
Hope that helps you
Hit the Thank You button as I am sharing my knowledge with you : )
Mark me as brainliest answer