Grade 12th passMechanics
A body starts from rest and moves with accerlation 5 m/s what is the distance travelled by abody in 4th second
3 Answers
sahil
sol:formula for distance covered in nth second is Sn= u + a/2(2n-1),Now here n is 4 and a is 5m and u is 0 so when we put S4=5m(2×4-1)÷2=35m/2.So the required answer is 35m/2 .If you found it helpful please approve the answer
Sanju
Use the eqn s = ut + ½ at2
Now substitute the values u = 0, a = 5 m/s2 and t = 4s in the above eqn.
Therefore, distance travelled by the body in 4th sec, s = (0*4) + (½ * 5 * 42) = 40 m
Shobhit
u or initial velocity is zero. Acceleration is 5ms/s^2. Therefore by,s=ut+1/2at^2 We have:s=0xt +1/2 *5*4*4s=40 m. Hence the distance or the displacement is 40m.
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