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A car starts its journey from the origin with the velocity 20i m/s and moves in x-y plane with acceleration of (2i + 4tj) m/s2. Then the speed of car at t = 2 s will be |
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Answer» vi is 20 m/s t is 2 s now to find a so draw a right angle triangle taking 2 at base and 4 at perpendicular..by finding hypoteneus which is resultant acceleration which is 4.4 now we have ALL values...putting in equation Vf=Vi+at vf =20 + 4.4x 2 vf= 28.2 Given u = 20 a = \(2\hat i+4\hat j\) |a| = \(\sqrt{4+16}\) |a| = \(\sqrt{20}\) v = u + at v = 20 + \(\sqrt{20}\) x 2 v = 20 + 8.82 v = 28.82 m/s |
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