plz answer this as soon as possible
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3m is the correct answer.
SresthaAbhi:
n this is a site to help each other..so its always better to elaborate ur answer
Answered by
1
Hi Arnav!!
As we know that displacement is equal to the area under the velocity - time graph, so first we need to find the area under this graph.
= area of the triangle AOB - area of the triangle BCD + area of the square DEFG
=(1/2 X 3 X 2) - (1/2 X 1 X 2) + (1X1)
= 3 - 1 + 1
= 3 m
Hence, the displacement of the body in 5 secs will be 3m
HOPE THAT CLEARS YOUR DOUBT FRIEND!!
PLEASE MARK AS BRAINLIESTT!!
As we know that displacement is equal to the area under the velocity - time graph, so first we need to find the area under this graph.
= area of the triangle AOB - area of the triangle BCD + area of the square DEFG
=(1/2 X 3 X 2) - (1/2 X 1 X 2) + (1X1)
= 3 - 1 + 1
= 3 m
Hence, the displacement of the body in 5 secs will be 3m
HOPE THAT CLEARS YOUR DOUBT FRIEND!!
PLEASE MARK AS BRAINLIESTT!!
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