An object starting from rest travels along a straight path with an acceleration of 5m/s2. What will be its velocity after 4s and calculate the distance covered by the object during this time? *

An object starting from rest travels along a straight path with an acc

1.	An object starting from rest travels along a straight path with an acceleration of 5m/s2. What will be its velocity after 4s and calculate the distance covered by the object during this time? *
Answer» ut+ 21 at 2 In first 2 s,s=20m20= 21 a(2) 2 =2aa=10 m/s 2 VELOCITY at the end of 2 s is v=u+at=0+10×2=20m/sIn next 4S, s=ut+ 21 a ′ t 2 160=20×4+ 21 a ′ (4) 2 80=8a ′ a ′ =10 m/s 2 It shows that acceleration is UNIFORM. From v=u+atv=0+10×7=70 m/s

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An object starting from rest travels along a straight path with an acceleration of 5m/s2. What will be its velocity after 4s and calculate the distance covered by the object during this time? *​

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An object starting from rest travels along a straight path with an acceleration of 5m/s2. What will be its velocity after 4s and calculate the distance covered by the object during this time? *