Thatstudentm9
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I feel embarrassed posting this but I don’t know how to do this
question
Application of calculus Mcq (1 Viewer)
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HoldingOn
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Displacement is represented by the area under the velocity curve. So we just calculate the area of the triangle we see i.e. 1/2 * 4 * 8= 16. But the particle starts at 2 so we have 16+2=18. Thus DI feel embarrassed posting this but I don’t know how to do thisquestion
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weaknuclearforce
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To see the "displacement under the curve" explicitly you can observe
integrating wrt t
=-2t+v_0 \ \ v(t)=-2t+8)
integrating wrt t
=-t^2+8t+x_0 \ \ x(t)=-t^2+8t+2)
You can see from the graph that maximum displacement must occur when velocity changes sign (direction) at t=4, or...
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integrating wrt t
integrating wrt t
You can see from the graph that maximum displacement must occur when velocity changes sign (direction) at t=4, or...