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<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_DPP01_DPP4.6_E01_345_Q01.png" width="80%"> <br> The acceleration time graph of a body is shown below the most probable velocity time graph of the body is

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Answer Text

<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_DPP01_DPP4.6_E01_345_O01.png" width="30%"><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_DPP01_DPP4.6_E01_345_O02.png" width="30%"><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_DPP01_DPP4.6_E01_345_O03.png" width="30%"><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_DPP01_DPP4.6_E01_345_O04.png" width="30%">

Answer :
C
Solution :
From given a-t graph it is clear that acceleration is increasing at constant rate <br> `(da)/(dt)=k` (constant)`impliesa=kt`(by integration) <br> `implies(dv)/(dt)=ktimpliesdvktdt` <br> `impliesintdv=kinttdtimpliesv=(kt^2)/(2)` <br> i.e., v is dependent on time parabolically and parabola is symmetric about v-axis. And suddenly acceleration becomes zero, i.e., velocity becomes constant. <br> Hence (c ) is most probable graph.

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