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An objects velocity changes at a constant rate from 2.0 m/s to 8.0 m/s in 5.0 s. The area underneath the acceleration vs. ti
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Step by step solution are as follows Given that Velocity Win = 2 omylec Velocrty Vt = 8.0 m/see- 4 time. t = 5.0 de Salutton a=flt; ladt gf 9=44) V aradt as an * Area enclosed between the acceleration-time cuive and - the part of time- aniß under consideration always gives change in vel

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